转自kuangbin的AC自动机（赛前最后一博）
2024-06-04 09:18:44
有了KMP和Trie的基础，就可以学习神奇的AC自动机了。AC自动机其实就是在Trie树上实现KMP，可以完成多模式串的匹配。
AC自动机 其实 就是创建了一个状态的转移图，思想很重要。
推荐的学习链接:
http://acm.uestc.edu.cn/bbs/read.php?tid=4294
http://blog.csdn.net/niushuai666/article/details/7002823
http://hi.baidu.com/nialv7/item/ce1ce015d44a6ba7feded52d
AC自动机专题训练链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=25605#overview 这里我提交的代码是公开的，可以看到
题目来源：http://www.notonlysuccess.com/index.php/aho-corasick-automaton/
写AC自动机的代码风格是向昀神学的，好简洁，写起来好棒的感觉。
1、HDU 2222 Keywords Search 最基本的入门题了
就是求目标串中出现了几个模式串。
很基础了。使用一个int型的end数组记录，查询一次。
//======================
// HDU 2222
// 求目标串中出现了几个模式串
//====================
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;struct Trie
{int next[500010][26],fail[500010],end[500010];int root,L;int newnode(){for(int i = 0;i < 26;i++)next[L][i] = -1;end[L++] = 0;return L-1;}void init(){L = 0;root = newnode();}void insert(char buf[]){int len = strlen(buf);int now = root;for(int i = 0;i < len;i++){if(next[now][buf[i]-'a'] == -1)next[now][buf[i]-'a'] = newnode();now = next[now][buf[i]-'a'];}end[now]++;}void build(){queue<int>Q;fail[root] = root;for(int i = 0;i < 26;i++)if(next[root][i] == -1)next[root][i] = root;else{fail[next[root][i]] = root;Q.push(next[root][i]);}while( !Q.empty() ){int now = Q.front();Q.pop();for(int i = 0;i < 26;i++)if(next[now][i] == -1)next[now][i] = next[fail[now]][i];else{fail[next[now][i]]=next[fail[now]][i];Q.push(next[now][i]);}}}int query(char buf[]){int len = strlen(buf);int now = root;int res = 0;for(int i = 0;i < len;i++){now = next[now][buf[i]-'a'];int temp = now;while( temp != root ){res += end[temp];end[temp] = 0;temp = fail[temp];}}return res;}void debug(){for(int i = 0;i < L;i++){printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);for(int j = 0;j < 26;j++)printf("%2d",next[i][j]);printf("]\n");}}
};
char buf[1000010];
Trie ac;
int main()
{int T;int n;scanf("%d",&T);while( T-- ){scanf("%d",&n);ac.init();for(int i = 0;i < n;i++){scanf("%s",buf);ac.insert(buf);}ac.build();scanf("%s",buf);printf("%d\n",ac.query(buf));}return 0;
}2、HDU 2896 病毒侵袭
这题和上题差不多，要输出出现模式串的id,用end记录id就可以了。还有trie树的分支是128的
题解here
//============================================================================
// Name : HDU.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;struct Trie
{int next[210*500][128],fail[210*500],end[210*500];int root,L;int newnode(){for(int i = 0;i < 128;i++)next[L][i] = -1;end[L++] = -1;return L-1;}void init(){L = 0;root = newnode();}void insert(char s[],int id){int len = strlen(s);int now = root;for(int i = 0;i < len;i++){if(next[now][s[i]] == -1)next[now][s[i]] = newnode();now=next[now][s[i]];}end[now]=id;}void build(){queue<int>Q;fail[root] = root;for(int i = 0;i < 128;i++)if(next[root][i] == -1)next[root][i] = root;else{fail[next[root][i]] = root;Q.push(next[root][i]);}while(!Q.empty()){int now = Q.front();Q.pop();for(int i = 0;i < 128;i++)if(next[now][i] == -1)next[now][i] = next[fail[now]][i];else{fail[next[now][i]] = next[fail[now]][i];Q.push(next[now][i]);}}}bool used[510];bool query(char buf[],int n,int id){int len = strlen(buf);int now = root;memset(used,false,sizeof(used));bool flag = false;for(int i = 0;i < len;i++){now = next[now][buf[i]];int temp = now;while(temp != root){if(end[temp] != -1){used[end[temp]] = true;flag = true;}temp = fail[temp];}}if(!flag)return false;printf("web %d:",id);for(int i = 1;i <= n;i++)if(used[i])printf(" %d",i);printf("\n");return true;}
};
char buf[10010];
Trie ac;
int main()
{int n,m;while(scanf("%d",&n) != EOF){ac.init();for(int i = 1;i <= n;i++){scanf("%s",buf);ac.insert(buf,i);}ac.build();int ans = 0;scanf("%d",&m);for(int i = 1;i <= m;i++){scanf("%s",buf);if(ac.query(buf,n,i))ans++;}printf("total: %d\n",ans);}return 0;
}View Code
3、HDU 3065 病毒侵袭持续中
这题的变化也不大，就是需要输出每个模式串出现的次数，查询的时候使用一个数组进行记录就可以了
1 //============================================================================
2 // Name : HDU.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <stdio.h>
11 #include <string.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15
16 char str[1010][100];
17 struct Trie
18 {
19 int next[1010*50][128],fail[1010*50],end[1010*50];
20 int root,L;
21 int newnode()
22 {
23 for(int i = 0;i < 128;i++)
24 next[L][i] = -1;
25 end[L++] = -1;
26 return L-1;
27 }
28 void init()
29 {
30 L = 0;
31 root = newnode();
32 }
33 void insert(char s[],int id)
34 {
35 int len = strlen(s);
36 int now = root;
37 for(int i = 0;i < len;i++)
38 {
39 if(next[now][s[i]] == -1)
40 next[now][s[i]] = newnode();
41 now = next[now][s[i]];
42 }
43 end[now] = id;
44 }
45 void build()
46 {
47 queue<int>Q;
48 fail[root] = root;
49 for(int i = 0;i < 128;i++)
50 if(next[root][i] == -1)
51 next[root][i] = root;
52 else
53 {
54 fail[next[root][i]] = root;
55 Q.push(next[root][i]);
56 }
57 while(!Q.empty())
58 {
59 int now = Q.front();
60 Q.pop();
61 for(int i = 0;i < 128;i++)
62 if(next[now][i] == -1)
63 next[now][i]=next[fail[now]][i];
64 else
65 {
66 fail[next[now][i]]=next[fail[now]][i];
67 Q.push(next[now][i]);
68 }
69 }
70 }
71 int num[1010];
72 void query(char buf[],int n)
73 {
74 for(int i = 0;i < n;i++)
75 num[i] = 0;
76 int len=strlen(buf);
77 int now=root;
78 for(int i=0;i<len;i++)
79 {
80 now=next[now][buf[i]];
81 int temp = now;
82 while( temp != root )
83 {
84 if(end[temp] != -1)
85 num[end[temp]]++;
86 temp = fail[temp];
87 }
88 }
89 for(int i = 0;i < n;i++)
90 if(num[i] > 0)
91 printf("%s: %d\n",str[i],num[i]);
92 }
93
94 };
95
96 char buf[2000010];
97 Trie ac;
98 void debug()
99 {
100 for (int i = 0; i < ac.L; i++)
101 {
102 printf("id = %3d ,fail = %3d ,end = %3d, chi = [",i,ac.fail[i],ac.end[i]);
103 for (int j = 0; j < 128; j++)
104 printf("%2d ",ac.next[i][j]);
105 printf("]\n");
106 }
107 }
108 int main()
109 {
110 // freopen("in.txt","r",stdin);
111 // freopen("out.txt","w",stdout);
112 int n;
113 while(scanf("%d",&n) == 1)
114 {
115 ac.init();
116 for(int i = 0;i < n;i++)
117 {
118 scanf("%s",str[i]);
119 ac.insert(str[i],i);
120 }
121 ac.build();
122 scanf("%s",buf);
123 ac.query(buf,n);
124 }
125 return 0;
126 }View Code
4、ZOJ 3430 Detect the Virus
主要是解码过程，解码以后就是模板题了。
求出现的模式串的种类数
分支需要256个
1 //============================================================================
2 // Name : ZOJ.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <stdio.h>
11 #include <string.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15
16 struct Trie
17 {
18 int next[520*64][256],fail[520*64],end[520*64];
19 int root,L;
20 int newnode()
21 {
22 for(int i = 0;i < 256;i++)
23 next[L][i] = -1;
24 end[L++] = -1;
25 return L-1;
26 }
27 void init()
28 {
29 L = 0;
30 root = newnode();
31 }
32 void insert(unsigned char buf[],int len,int id)
33 {
34 int now = root;
35 for(int i = 0;i < len;i++)
36 {
37 if(next[now][buf[i]] == -1)
38 next[now][buf[i]] = newnode();
39 now = next[now][buf[i]];
40 }
41 end[now] = id;
42 }
43 void build()
44 {
45 queue<int>Q;
46 fail[root] = root;
47 for(int i = 0;i < 256;i++)
48 if(next[root][i] == -1)
49 next[root][i] = root;
50 else
51 {
52 fail[next[root][i]]=root;
53 Q.push(next[root][i]);
54 }
55 while(!Q.empty())
56 {
57 int now = Q.front();
58 Q.pop();
59 for(int i = 0;i < 256;i++)
60 if(next[now][i] == -1)
61 next[now][i] = next[fail[now]][i];
62 else
63 {
64 fail[next[now][i]] = next[fail[now]][i];
65 Q.push(next[now][i]);
66 }
67 }
68 }
69 bool used[520];
70 int query(unsigned char buf[],int len,int n)
71 {
72 memset(used,false,sizeof(used));
73 int now = root;
74 for(int i = 0;i < len;i++)
75 {
76 now = next[now][buf[i]];
77 int temp = now;
78 while( temp!=root )
79 {
80 if(end[temp] != -1)
81 used[end[temp]]=true;
82 temp = fail[temp];
83 }
84 }
85 int res = 0;
86 for(int i = 0;i < n;i++)
87 if(used[i])
88 res++;
89 return res;
90 }
91 };
92
93 unsigned char buf[2050];
94 int tot;
95 char str[4000];
96 unsigned char s[4000];
97 unsigned char Get(char ch)
98 {
99 if( ch>='A'&&ch<='Z' )return ch-'A';
100 if( ch>='a'&&ch<='z' )return ch-'a'+26;
101 if( ch>='0'&&ch<='9' )return ch-'0'+52;
102 if( ch=='+' )return 62;
103 else return 63;
104 }
105 void change(unsigned char str[],int len)
106 {
107 int t=0;
108 for(int i=0;i<len;i+=4)
109 {
110 buf[t++]=((str[i]<<2)|(str[i+1]>>4));
111 if(i+2 < len)
112 buf[t++]=( (str[i+1]<<4)|(str[i+2]>>2) );
113 if(i+3 < len)
114 buf[t++]= ( (str[i+2]<<6)|str[i+3] );
115 }
116 tot=t;
117 }
118 Trie ac;
119 int main()
120 {
121 // freopen("in.txt","r",stdin);
122 // freopen("out.txt","w",stdout);
123 int n,m;
124 while(scanf("%d",&n) == 1)
125 {
126 ac.init();
127 for(int i = 0;i < n;i++)
128 {
129 scanf("%s",str);
130 int len = strlen(str);
131 while(str[len-1]=='=')len--;
132 for(int j = 0;j < len;j++)
133 {
134 s[j] = Get(str[j]);
135 }
136 change(s,len);
137 ac.insert(buf,tot,i);
138 }
139 ac.build();
140 scanf("%d",&m);
141 while(m--)
142 {
143 scanf("%s",str);
144 int len=strlen(str);
145 while(str[len-1]=='=')len--;
146 for(int j = 0;j < len;j++)
147 s[j] = Get(str[j]);
148 change(s,len);
149 printf("%d\n",ac.query(buf,tot,n));
150 }
151 printf("\n");
152 }
153 return 0;
154 }View Code
5、POJ 2778 DNA Sequence
AC自动机+矩阵加速
这个时候AC自动机 的一种状态转移图的思路就很透彻了。
AC自动机就是可以确定状态的转移。
1 //============================================================================
2 // Name : POJ.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <stdio.h>
11 #include <algorithm>
12 #include <string.h>
13 #include <queue>
14 using namespace std;
15
16 const int MOD=100000;
17 struct Matrix
18 {
19 int mat[110][110],n;
20 Matrix(){}
21 Matrix(int _n)
22 {
23 n = _n;
24 for(int i=0;i<n;i++)
25 for(int j=0;j<n;j++)
26 mat[i][j]=0;
27 }
28 Matrix operator *(const Matrix &b)const
29 {
30 Matrix ret=Matrix(n);
31 for(int i=0;i<n;i++)
32 for(int j=0;j<n;j++)
33 for(int k=0;k<n;k++)
34 {
35 int tmp=(long long)mat[i][k]*b.mat[k][j]%MOD;
36 ret.mat[i][j]=(ret.mat[i][j]+tmp)%MOD;
37 }
38 return ret;
39 }
40 };
41 struct Trie
42 {
43 int next[110][4],fail[110];
44 bool end[110];
45 int root,L;
46 int newnode()
47 {
48 for(int i=0;i<4;i++)
49 next[L][i]=-1;
50 end[L++]=false;
51 return L-1;
52 }
53 void init()
54 {
55 L=0;
56 root=newnode();
57 }
58 int getch(char ch)
59 {
60 switch(ch)
61 {
62 case 'A':
63 return 0;
64 break;
65 case 'C':
66 return 1;
67 break;
68 case 'G':
69 return 2;
70 break;
71 case 'T':
72 return 3;
73 break;
74 }
75 }
76 void insert(char s[])
77 {
78 int len=strlen(s);
79 int now=root;
80 for(int i = 0;i < len;i++)
81 {
82 if(next[now][getch(s[i])] == -1)
83 next[now][getch(s[i])] = newnode();
84 now = next[now][getch(s[i])];
85 }
86 end[now]=true;
87 }
88 void build()
89 {
90 queue<int>Q;
91 for(int i = 0;i < 4;i++)
92 if(next[root][i] == -1)
93 next[root][i] = root;
94 else
95 {
96 fail[next[root][i]] = root;
97 Q.push(next[root][i]);
98 }
99 while(!Q.empty())
100 {
101 int now = Q.front();
102 Q.pop();
103 if(end[fail[now]]==true)
104 end[now]=true;
105 for(int i = 0;i < 4;i++)
106 {
107 if(next[now][i] == -1)
108 next[now][i] = next[fail[now]][i];
109 else
110 {
111 fail[next[now][i]] = next[fail[now]][i];
112 Q.push(next[now][i]);
113 }
114 }
115 }
116 }
117 Matrix getMatrix()
118 {
119 Matrix res = Matrix(L);
120 for(int i=0;i<L;i++)
121 for(int j=0;j<4;j++)
122 if(end[next[i][j]]==false)
123 res.mat[i][next[i][j]]++;
124 return res;
125 }
126 };
127
128 Trie ac;
129 char buf[20];
130
131 Matrix pow_M(Matrix a,int n)
132 {
133 Matrix ret = Matrix(a.n);
134 for(int i = 0; i < ret.n; i++)
135 ret.mat[i][i]=1;
136 Matrix tmp=a;
137 while(n)
138 {
139 if(n&1)ret=ret*tmp;
140 tmp=tmp*tmp;
141 n>>=1;
142 }
143 return ret;
144 }
145
146 int main()
147 {
148 int n,m;
149 while(scanf("%d%d",&n,&m) != EOF)
150 {
151 ac.init();
152 for(int i=0;i<n;i++)
153 {
154 scanf("%s",buf);
155 ac.insert(buf);
156 }
157 ac.build();
158 Matrix a=ac.getMatrix();
159 a=pow_M(a,m);
160 int ans=0;
161 for(int i=0;i<a.n;i++)
162 {
163 ans=(ans+a.mat[0][i])%MOD;
164 }
165 printf("%d\n",ans);
166 }
167 return 0;
168 }View Code
6、HDU 2243 考研路茫茫——单词情结
这题和上题有些类似。但是需要求和。
所以给
矩阵增加一维，这样可以完美解决
题解here
1 //============================================================================
2 // Name : HDU.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <stdio.h>
11 #include <string.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15 struct Matrix
16 {
17 unsigned long long mat[40][40];
18 int n;
19 Matrix(){}
20 Matrix(int _n)
21 {
22 n=_n;
23 for(int i=0;i<n;i++)
24 for(int j=0;j<n;j++)
25 mat[i][j] = 0;
26 }
27 Matrix operator *(const Matrix &b)const
28 {
29 Matrix ret = Matrix(n);
30 for(int i=0;i<n;i++)
31 for(int j=0;j<n;j++)
32 for(int k=0;k<n;k++)
33 ret.mat[i][j]+=mat[i][k]*b.mat[k][j];
34 return ret;
35 }
36 };
37 unsigned long long pow_m(unsigned long long a,int n)
38 {
39 unsigned long long ret=1;
40 unsigned long long tmp = a;
41 while(n)
42 {
43 if(n&1)ret*=tmp;
44 tmp*=tmp;
45 n>>=1;
46 }
47 return ret;
48 }
49 Matrix pow_M(Matrix a,int n)
50 {
51 Matrix ret = Matrix(a.n);
52 for(int i=0;i<a.n;i++)
53 ret.mat[i][i] = 1;
54 Matrix tmp = a;
55 while(n)
56 {
57 if(n&1)ret=ret*tmp;
58 tmp=tmp*tmp;
59 n>>=1;
60 }
61 return ret;
62 }
63 struct Trie
64 {
65 int next[40][26],fail[40];
66 bool end[40];
67 int root,L;
68 int newnode()
69 {
70 for(int i = 0;i < 26;i++)
71 next[L][i] = -1;
72 end[L++] = false;
73 return L-1;
74 }
75 void init()
76 {
77 L = 0;
78 root = newnode();
79 }
80 void insert(char buf[])
81 {
82 int len = strlen(buf);
83 int now = root;
84 for(int i = 0;i < len;i++)
85 {
86 if(next[now][buf[i]-'a'] == -1)
87 next[now][buf[i]-'a'] = newnode();
88 now = next[now][buf[i]-'a'];
89 }
90 end[now] = true;
91 }
92 void build()
93 {
94 queue<int>Q;
95 fail[root]=root;
96 for(int i = 0;i < 26;i++)
97 if(next[root][i] == -1)
98 next[root][i] = root;
99 else
100 {
101 fail[next[root][i]] = root;
102 Q.push(next[root][i]);
103 }
104 while(!Q.empty())
105 {
106 int now = Q.front();
107 Q.pop();
108 if(end[fail[now]])end[now]=true;
109 for(int i = 0;i < 26;i++)
110 if(next[now][i] == -1)
111 next[now][i] = next[fail[now]][i];
112 else
113 {
114 fail[next[now][i]] = next[fail[now]][i];
115 Q.push(next[now][i]);
116 }
117 }
118 }
119 Matrix getMatrix()
120 {
121 Matrix ret = Matrix(L+1);
122 for(int i = 0;i < L;i++)
123 for(int j = 0;j < 26;j++)
124 if(end[next[i][j]]==false)
125 ret.mat[i][next[i][j]] ++;
126 for(int i = 0;i < L+1;i++)
127 ret.mat[i][L] = 1;
128 return ret;
129 }
130 void debug()
131 {
132 for(int i = 0;i < L;i++)
133 {
134 printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
135 for(int j = 0;j < 26;j++)
136 printf("%2d",next[i][j]);
137 printf("]\n");
138 }
139 }
140 };
141 char buf[10];
142 Trie ac;
143 int main()
144 {
145 // freopen("in.txt","r",stdin);
146 // freopen("out.txt","w",stdout);
147 int n,L;
148 while(scanf("%d%d",&n,&L)==2)
149 {
150 ac.init();
151 for(int i = 0;i < n;i++)
152 {
153 scanf("%s",buf);
154 ac.insert(buf);
155 }
156 ac.build();
157 Matrix a = ac.getMatrix();
158 a = pow_M(a,L);
159 unsigned long long res = 0;
160 for(int i = 0;i < a.n;i++)
161 res += a.mat[0][i];
162 res--;
163
164 /*
165 * f[n]=1 + 26^1 + 26^2 +...26^n
166 * f[n]=26*f[n-1]+1
167 * {f[n] 1} = {f[n-1] 1}[26 0;1 1]
168 * 数是f[L]-1;
169 * 此题的L<2^31.矩阵的幂不能是L+1次，否则就超时了
170 */
171 a = Matrix(2);
172 a.mat[0][0]=26;
173 a.mat[1][0] = a.mat[1][1] = 1;
174 a=pow_M(a,L);
175 unsigned long long ans=a.mat[1][0]+a.mat[0][0];
176 ans--;
177 ans-=res;
178 cout<<ans<<endl;
179 }
180 return 0;
181 }View Code
7、POJ 1625 Censored!
AC自动机+DP+高精度
好题
题解here
//============================================================================
// Name : POJ.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <map>
using namespace std;
map<char,int>mp;
int N,M,P;
struct Matrix
{int mat[110][110];int n;Matrix(){}Matrix(int _n){n=_n;for(int i = 0;i < n;i++)for(int j = 0;j < n;j++)mat[i][j] = 0;}
};
struct Trie
{int next[110][256],fail[110];bool end[110];int L,root;int newnode(){for(int i = 0;i < 256;i++)next[L][i] = -1;end[L++] = false;return L-1;}void init(){L = 0;root = newnode();}void insert(char buf[]){int len = strlen(buf);int now = root;for(int i = 0;i < len;i++){if(next[now][mp[buf[i]]] == -1)next[now][mp[buf[i]]] = newnode();now = next[now][mp[buf[i]]];}end[now] = true;}void build(){queue<int>Q;fail[root] = root;for(int i = 0;i < 256;i++)if(next[root][i] == -1)next[root][i] = root;else{fail[next[root][i]] = root;Q.push(next[root][i]);}while(!Q.empty()){int now = Q.front();Q.pop();if(end[fail[now]]==true)end[now]=true;for(int i = 0;i < 256;i++)if(next[now][i] == -1)next[now][i] = next[fail[now]][i];else{fail[next[now][i]] = next[fail[now]][i];Q.push(next[now][i]);}}}Matrix getMatrix(){Matrix res = Matrix(L);for(int i = 0;i < L;i++)for(int j = 0;j < N;j++)if(end[next[i][j]]==false)res.mat[i][next[i][j]]++;return res;}void debug(){for(int i = 0;i < L;i++){printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);for(int j = 0;j < 26;j++)printf("%2d",next[i][j]);printf("]\n");}}};/** 高精度，支持乘法和加法*/
struct BigInt
{const static int mod = 10000;const static int DLEN = 4;int a[600],len;BigInt(){memset(a,0,sizeof(a));len = 1;}BigInt(int v){memset(a,0,sizeof(a));len = 0;do{a[len++] = v%mod;v /= mod;}while(v);}BigInt(const char s[]){memset(a,0,sizeof(a));int L = strlen(s);len = L/DLEN;if(L%DLEN)len++;int index = 0;for(int i = L-1;i >= 0;i -= DLEN){int t = 0;int k = i - DLEN + 1;if(k < 0)k = 0;for(int j = k;j <= i;j++)t = t*10 + s[j] - '0';a[index++] = t;}}BigInt operator +(const BigInt &b)const{BigInt res;res.len = max(len,b.len);for(int i = 0;i <= res.len;i++)res.a[i] = 0;for(int i = 0;i < res.len;i++){res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);res.a[i+1] += res.a[i]/mod;res.a[i] %= mod;}if(res.a[res.len] > 0)res.len++;return res;}BigInt operator *(const BigInt &b)const{BigInt res;for(int i = 0; i < len;i++){int up = 0;for(int j = 0;j < b.len;j++){int temp = a[i]*b.a[j] + res.a[i+j] + up;res.a[i+j] = temp%mod;up = temp/mod;}if(up != 0)res.a[i + b.len] = up;}res.len = len + b.len;while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;return res;}void output(){printf("%d",a[len-1]);for(int i = len-2;i >=0 ;i--)printf("%04d",a[i]);printf("\n");}
};
char buf[1010];
BigInt dp[2][110];
Trie ac;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);while(scanf("%d%d%d",&N,&M,&P)==3){gets(buf);gets(buf);mp.clear();int len = strlen(buf);for(int i = 0;i < len;i++)mp[buf[i]]=i;ac.init();for(int i = 0;i < P;i++){gets(buf);ac.insert(buf);}ac.build();
// ac.debug();Matrix a= ac.getMatrix();// for(int i = 0;i <a.n;i++)
// {
// for(int j=0;j<a.n;j++)printf("%d ",a.mat[i][j]);
// cout<<endl;
// }int now = 0;dp[now][0] = 1;for(int i = 1;i < a.n;i++)dp[now][i] = 0;for(int i = 0;i < M;i++){now^=1;for(int j = 0;j < a.n;j++)dp[now][j] = 0;for(int j = 0;j < a.n;j++)for(int k = 0;k < a.n;k++)if(a.mat[j][k] > 0)dp[now][k] = dp[now][k]+dp[now^1][j]*a.mat[j][k];
// for(int j = 0;j < a.n;j++)
// dp[now][j].output();
}BigInt ans = 0;for(int i = 0;i < a.n;i++)ans = ans + dp[now][i];ans.output();}return 0;
}View Code
8、HDU 2825 Wireless Password
AC自动机+状态压缩DP
相当于在AC自动机上产生状态转移，然后进行dp
//============================================================================
// Name : HDU.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
const int MOD=20090717;
int n,m,k;
int dp[30][110][1<<10];
int num[5000];struct Trie
{int next[110][26],fail[110],end[110];int root,L;int newnode(){for(int i = 0;i < 26;i++)next[L][i] = -1;end[L++] = 0;return L-1;}void init(){L = 0;root = newnode();}void insert(char buf[],int id){int len = strlen(buf);int now = root;for(int i = 0;i < len;i++){if(next[now][buf[i]-'a']==-1)next[now][buf[i]-'a'] = newnode();now = next[now][buf[i]-'a'];}end[now] |= (1<<id);}void build(){queue<int>Q;fail[root] = root;for(int i = 0;i < 26;i++)if(next[root][i] == -1)next[root][i] = root;else{fail[next[root][i]] = root;Q.push(next[root][i]);}while(!Q.empty()){int now = Q.front();Q.pop();end[now] |= end[fail[now]];for(int i = 0;i < 26;i++)if(next[now][i] == -1)next[now][i] = next[fail[now]][i];else{fail[next[now][i]] = next[fail[now]][i];Q.push(next[now][i]);}}}int solve(){//memset(dp,0,sizeof(dp));for(int i = 0;i <= n;i++)for(int j = 0;j < L;j++)for(int p = 0;p < (1<<m);p++)dp[i][j][p]=0;dp[0][0][0] = 1;for(int i = 0;i < n;i++)for(int j = 0;j < L;j++)for(int p = 0;p< (1<<m);p++)if(dp[i][j][p] > 0){for(int x = 0;x < 26;x++){int newi = i+1;int newj = next[j][x];int newp = (p|end[newj]);dp[newi][newj][newp] += dp[i][j][p];dp[newi][newj][newp]%=MOD;}}int ans = 0;for(int p = 0;p < (1<<m);p++){if(num[p] < k)continue;for(int i = 0;i < L;i++){ans = (ans + dp[n][i][p])%MOD;}}return ans;}
};
char buf[20];
Trie ac;
int main()
{for(int i=0;i<(1<<10);i++){num[i] = 0;for(int j = 0;j < 10;j++)if(i & (1<<j))num[i]++;}while(scanf("%d%d%d",&n,&m,&k)==3){if(n== 0 && m==0 &&k==0)break;ac.init();for(int i = 0;i < m;i++){scanf("%s",buf);ac.insert(buf,i);}ac.build();printf("%d\n",ac.solve());}return 0;
}View Code
9、HDU 2296 Ring
需要输出字典序最小的解
在DP的时候加一个字符数组来记录就行了
//============================================================================
// Name : HDU.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;int a[110];
int dp[55][1110];
char str[55][1110][55];bool cmp(char s1[],char s2[])
{int len1=strlen(s1);int len2=strlen(s2);if(len1 != len2)return len1 < len2;else return strcmp(s1,s2) < 0;
}const int INF=0x3f3f3f3f;
struct Trie
{int next[1110][26],fail[1110],end[1110];int root,L;int newnode(){for(int i = 0;i < 26;i++)next[L][i] = -1;end[L++] = -1;return L-1;}void init(){L = 0;root = newnode();}void insert(char buf[],int id){int len = strlen(buf);int now = root;for(int i = 0;i < len;i++){if(next[now][buf[i]-'a'] == -1)next[now][buf[i]-'a'] = newnode();now = next[now][buf[i]-'a'];}end[now] = id;}void build(){queue<int>Q;fail[root] = root;for(int i = 0;i < 26;i++)if(next[root][i] == -1)next[root][i] = root;else{fail[next[root][i]] = root;Q.push(next[root][i]);}while(!Q.empty()){int now = Q.front();Q.pop();for(int i = 0;i < 26;i++)if(next[now][i] == -1)next[now][i] = next[fail[now]][i];else{fail[next[now][i]] = next[fail[now]][i];Q.push(next[now][i]);}}}void solve(int n){for(int i = 0;i <= n;i++)for(int j = 0;j < L;j++)dp[i][j] = -INF;dp[0][0] = 0;strcpy(str[0][0],"");char ans[55];strcpy(ans,"");int Max = 0;char tmp[55];for(int i = 0; i < n;i++)for(int j = 0;j < L;j++)if(dp[i][j]>=0){strcpy(tmp,str[i][j]);int len = strlen(tmp);for(int k = 0;k < 26;k++){int nxt=next[j][k];tmp[len] = 'a'+k;tmp[len+1] = 0;int tt = dp[i][j];if(end[nxt] != -1)tt+=a[end[nxt]];if(dp[i+1][nxt]<tt || (dp[i+1][nxt]==tt && cmp(tmp,str[i+1][nxt]))){dp[i+1][nxt] = tt;strcpy(str[i+1][nxt],tmp);if(tt > Max ||(tt==Max && cmp(tmp,ans))){Max = tt;strcpy(ans,tmp);}}}}printf("%s\n",ans);}
};
char buf[60];
Trie ac;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);int T;int n,m;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);ac.init();for(int i = 0;i < m;i++){scanf("%s",buf);ac.insert(buf,i);}for(int i = 0;i < m;i++)scanf("%d",&a[i]);ac.build();ac.solve(n);}return 0;
}View Code
10、HDU 2457 DNA repair
很简单的AC自动机+DP了
1 //============================================================================
2 // Name : HDU.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <string.h>
11 #include <stdio.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15 const int INF = 0x3f3f3f3f;
16 struct Trie
17 {
18 int next[1010][4],fail[1010];
19 bool end[1010];
20 int root,L;
21 int newnode()
22 {
23 for(int i = 0;i < 4;i++)
24 next[L][i] = -1;
25 end[L++] = false;
26 return L-1;
27 }
28 void init()
29 {
30 L = 0;
31 root = newnode();
32 }
33 int getch(char ch)
34 {
35 if(ch == 'A')return 0;
36 else if(ch == 'C')return 1;
37 else if(ch == 'G')return 2;
38 else if(ch == 'T')return 3;
39 }
40 void insert(char buf[])
41 {
42 int len = strlen(buf);
43 int now = root;
44 for(int i = 0;i < len;i++)
45 {
46 if(next[now][getch(buf[i])] == -1)
47 next[now][getch(buf[i])] = newnode();
48 now = next[now][getch(buf[i])];
49 }
50 end[now] = true;
51 }
52 void build()
53 {
54 queue<int>Q;
55 fail[root] = root;
56 for(int i = 0;i < 4;i++)
57 if(next[root][i] == -1)
58 next[root][i] = root;
59 else
60 {
61 fail[next[root][i]] = root;
62 Q.push(next[root][i]);
63 }
64 while(!Q.empty())
65 {
66 int now = Q.front();
67 Q.pop();
68 if(end[fail[now]])end[now] = true;//这里不要忘记
69 for(int i = 0;i < 4;i++)
70 if(next[now][i] == -1)
71 next[now][i] = next[fail[now]][i];
72 else
73 {
74 fail[next[now][i]] = next[fail[now]][i];
75 Q.push(next[now][i]);
76 }
77 }
78 }
79 int dp[1010][1010];
80 int solve(char buf[])
81 {
82 int len = strlen(buf);
83 for(int i = 0;i <= len;i++)
84 for(int j = 0;j < L;j++)
85 dp[i][j] = INF;
86 dp[0][root] = 0;
87 for(int i = 0;i < len;i++)
88 for(int j = 0;j < L;j++)
89 if(dp[i][j] < INF)
90 {
91 for(int k = 0;k < 4;k++)
92 {
93 int news = next[j][k];
94 if(end[news])continue;
95 int tmp;
96 if( k == getch(buf[i]))tmp = dp[i][j];
97 else tmp = dp[i][j] + 1;
98 dp[i+1][news] = min(dp[i+1][news],tmp);
99 }
100 }
101 int ans = INF;
102 for(int j = 0;j < L;j++)
103 ans = min(ans,dp[len][j]);
104 if(ans == INF)ans = -1;
105 return ans;
106 }
107
108 };
109 char buf[1010];
110 Trie ac;
111 int main()
112 {
113 int n;
114 int iCase = 0;
115 while ( scanf("%d",&n) == 1 && n)
116 {
117 iCase++;
118 ac.init();
119 while(n--)
120 {
121 scanf("%s",buf);
122 ac.insert(buf);
123 }
124 ac.build();
125 scanf("%s",buf);
126 printf("Case %d: %d\n",iCase,ac.solve(buf));
127 }
128 return 0;
129 }View Code
11、ZOJ 3228 Searching the String
这题需要查询两种，一种是可重叠，一种是不可重叠的。
找模式串在目标串中的出现次数。
加一个数组记录上一次出现的位置，然后就可以求出不可重叠的了
1 //============================================================================
2 // Name : ZOJ.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <stdio.h>
11 #include <string.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15
16 struct Trie
17 {
18 int next[600010][26],fail[600010],deep[600010];
19 int root,L;
20 int newnode()
21 {
22 for(int i = 0;i < 26;i++)
23 next[L][i] = -1;
24 L++;
25 return L-1;
26 }
27 void init()
28 {
29 L = 0;
30 root = newnode();
31 deep[root] = 0;
32 }
33 int insert(char buf[])
34 {
35 int len = strlen(buf);
36 int now = root;
37 for(int i = 0;i < len;i++)
38 {
39 if(next[now][buf[i]-'a'] == -1)
40 {
41 next[now][buf[i]-'a'] = newnode();
42 deep[ next[now][buf[i]-'a'] ] = i+1;
43 }
44 now = next[now][buf[i]-'a'];
45 }
46 return now;
47 }
48 void build()
49 {
50 queue<int>Q;
51 fail[root] = root;
52 for(int i = 0;i < 26;i++)
53 if(next[root][i] == -1)
54 next[root][i] = root;
55 else
56 {
57 fail[next[root][i]] = root;
58 Q.push(next[root][i]);
59 }
60 while(!Q.empty())
61 {
62 int now = Q.front();
63 Q.pop();
64 for(int i = 0;i < 26;i++)
65 if(next[now][i] == -1)
66 next[now][i] = next[fail[now]][i];
67 else
68 {
69 fail[next[now][i]] = next[fail[now]][i];
70 Q.push(next[now][i]);
71 }
72 }
73 }
74 int cnt[600010][2];
75 int last[600010];
76 void query(char buf[])
77 {
78 int len = strlen(buf);
79 int now = root;
80 memset(cnt,0,sizeof(cnt));
81 memset(last,-1,sizeof(last));
82 for(int i = 0;i < len;i++)
83 {
84 now = next[now][buf[i]-'a'];
85 int temp = now;
86 while(temp != root)
87 {
88 cnt[temp][0]++;
89 if(i - last[temp] >= deep[temp])
90 {
91 last[temp] = i;
92 cnt[temp][1]++;
93 }
94 temp = fail[temp];
95 }
96 }
97 }
98 };
99 Trie ac;
100 char str[100010];
101 char buf[20];
102 int typ[100010],pos[100010];
103 int main()
104 {
105 // freopen("in.txt","r",stdin);
106 // freopen("out.txt","w",stdout);
107 int n;
108 int iCase = 0;
109 while(scanf("%s",str)==1)
110 {
111 iCase++;
112 printf("Case %d\n",iCase);
113 scanf("%d",&n);
114 ac.init();
115 for(int i = 0;i < n;i++)
116 {
117 scanf("%d%s",&typ[i],buf);
118 pos[i]=ac.insert(buf);
119 }
120 ac.build();
121 ac.query(str);
122 for(int i = 0;i < n;i++)
123 printf("%d\n",ac.cnt[pos[i]][typ[i]]);
124 printf("\n");
125 }
126 return 0;
127 }View Code
12、HDU 3341 Lost's revenge
这题主要是状态的表示，就是记录ACGT出现的次数。
然后这个ACGT次数表示的时候，状态要转化。
题解here
1 //============================================================================
2 // Name : HDU.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <string.h>
11 #include <stdio.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15 const int INF = 0x3f3f3f3f;
16 struct Trie
17 {
18 int next[510][4],fail[510];
19 int end[510];
20 int root,L;
21 int newnode()
22 {
23 for(int i = 0;i < 4;i++)
24 next[L][i] = -1;
25 end[L++] = 0;
26 return L-1;
27 }
28 void init()
29 {
30 L = 0;
31 root = newnode();
32 }
33 int getch(char ch)
34 {
35 if(ch == 'A')return 0;
36 else if(ch == 'C')return 1;
37 else if(ch == 'G')return 2;
38 else return 3;
39 }
40 void insert(char buf[])
41 {
42 int len = strlen(buf);
43 int now = root;
44 for(int i = 0;i < len;i++)
45 {
46 if(next[now][getch(buf[i])] == -1)
47 next[now][getch(buf[i])] = newnode();
48 now = next[now][getch(buf[i])];
49 }
50 end[now] ++;
51 }
52 void build()
53 {
54 queue<int>Q;
55 fail[root] = root;
56 for(int i = 0;i < 4;i++)
57 if(next[root][i] == -1)
58 next[root][i] = root;
59 else
60 {
61 fail[next[root][i]] = root;
62 Q.push(next[root][i]);
63 }
64 while(!Q.empty())
65 {
66 int now = Q.front();
67 Q.pop();
68 end[now] += end[fail[now]];/********/
69 for(int i = 0;i < 4;i++)
70 if(next[now][i] == -1)
71 next[now][i] = next[fail[now]][i];
72 else
73 {
74 fail[next[now][i]] = next[fail[now]][i];
75 Q.push(next[now][i]);
76 }
77 }
78 }
79 int dp[510][11*11*11*11+5];
80 int bit[4];
81 int num[4];
82 int solve(char buf[])
83 {
84 int len = strlen(buf);
85 memset(num,0,sizeof(num));
86 for(int i = 0;i < len;i++)
87 num[getch(buf[i])]++;
88 bit[0] = (num[1]+1)*(num[2]+1)*(num[3]+1);
89 bit[1] = (num[2]+1)*(num[3]+1);
90 bit[2] = (num[3]+1);
91 bit[3] = 1;
92 memset(dp,-1,sizeof(dp));
93 dp[root][0] = 0;
94 for(int A = 0;A <= num[0];A++)
95 for(int B = 0;B <= num[1];B++)
96 for(int C = 0;C <= num[2];C++)
97 for(int D = 0;D <= num[3];D++)
98 {
99 int s = A*bit[0] + B*bit[1] + C*bit[2] + D*bit[3];
100 for(int i = 0;i < L;i++)
101 if(dp[i][s] >= 0)
102 {
103 for(int k = 0;k < 4;k++)
104 {
105 if(k == 0 && A == num[0])continue;
106 if(k == 1 && B == num[1])continue;
107 if(k == 2 && C == num[2])continue;
108 if(k == 3 && D == num[3])continue;
109 dp[next[i][k]][s+bit[k]] = max(dp[next[i][k]][s+bit[k]],dp[i][s]+end[next[i][k]]);
110 }
111 }
112 }
113 int ans = 0;
114 int status = num[0]*bit[0] + num[1]*bit[1] + num[2]*bit[2] + num[3]*bit[3];
115 for(int i = 0;i < L;i++)
116 ans = max(ans,dp[i][status]);
117 return ans;
118 }
119 };
120 char buf[50];
121 Trie ac;
122 int main()
123 {
124 // freopen("in.txt","r",stdin);
125 // freopen("out.txt","w",stdout);
126 int n;
127 int iCase = 0;
128 while( scanf("%d",&n) == 1 && n)
129 {
130 iCase++;
131 ac.init();
132 for(int i = 0;i < n;i++)
133 {
134 scanf("%s",buf);
135 ac.insert(buf);
136 }
137 ac.build();
138 scanf("%s",buf);
139 printf("Case %d: %d\n",iCase,ac.solve(buf));
140 }
141 return 0;
142 }View Code
13、HDU 3247 Resource Archiver
使用最短路预处理出状态的转移。这样可以优化很多
1 //============================================================================
2 // Name : HDU.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <stdio.h>
11 #include <string.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15
16 const int INF = 0x3f3f3f3f;
17
18 struct Trie
19 {
20 int next[60010][2],fail[60010],end[60010];
21 int root,L;
22 int newnode()
23 {
24 for(int i = 0;i < 2;i++)
25 next[L][i] = -1;
26 end[L++] = 0;
27 return L-1;
28 }
29 void init()
30 {
31 L = 0;
32 root = newnode();
33 }
34 void insert(char buf[],int id)
35 {
36 int len = strlen(buf);
37 int now = root;
38 for(int i = 0;i < len ;i++)
39 {
40 if(next[now][buf[i]-'0'] == -1)
41 next[now][buf[i]-'0'] = newnode();
42 now = next[now][buf[i]-'0'];
43 }
44 end[now] = id;
45 }
46 void build()
47 {
48 queue<int>Q;
49 fail[root] = root;
50 for(int i = 0;i < 2;i++)
51 if(next[root][i] == -1)
52 next[root][i] = root;
53 else
54 {
55 fail[next[root][i]] = root;
56 Q.push(next[root][i]);
57 }
58 while( !Q.empty() )
59 {
60 int now = Q.front();
61 Q.pop();
62 if(end[fail[now]] == -1)end[now] = -1;
63 else end[now] |= end[fail[now]];
64 for(int i = 0;i < 2;i++)
65 if(next[now][i] == -1)
66 next[now][i] = next[fail[now]][i];
67 else
68 {
69 fail[next[now][i]] = next[fail[now]][i];
70 Q.push(next[now][i]);
71 }
72 }
73 }
74 int g[11][11];
75 int dp[1025][11];
76 int cnt;
77 int pos[11];
78 int dis[60010];
79
80
81 void bfs(int k)
82 {
83 queue<int>q;
84 memset(dis,-1,sizeof(dis));
85 dis[pos[k]] = 0;
86 q.push(pos[k]);
87 while(!q.empty())
88 {
89 int now = q.front();
90 q.pop();
91 for(int i = 0; i< 2;i++)
92 {
93 int tmp = next[now][i];
94 if(dis[tmp]<0 && end[tmp] >= 0)
95 {
96 dis[tmp] = dis[now] + 1;
97 q.push(tmp);
98 }
99 }
100 }
101 for(int i = 0;i < cnt;i++)
102 g[k][i] = dis[pos[i]];
103 }
104
105
106 int solve(int n)
107 {
108
109 pos[0] = 0;
110 cnt = 1;
111 for(int i = 0;i < L;i++)
112 if(end[i] > 0)
113 pos[cnt++] = i;
114 for(int i = 0; i < cnt;i++)
115 bfs(i);
116
117 for(int i = 0;i < (1<<n);i++)
118 for(int j = 0;j < cnt;j++)
119 dp[i][j] = INF;
120 dp[0][0] = 0;
121 for(int i = 0;i <(1<<n);i++)
122 for(int j = 0;j < cnt;j++)
123 if(dp[i][j]<INF)
124 {
125 for(int k = 0;k < cnt;k++)
126 {
127 if(g[j][k] < 0)continue;
128 if( j == k)continue;
129 dp[i|end[pos[k]]][k] = min(dp[i|end[pos[k]]][k],dp[i][j]+g[j][k]);
130 }
131 }
132 int ans = INF;
133 for(int j = 0;j < cnt;j++)
134 ans = min(ans,dp[(1<<n)-1][j]);
135 return ans;
136 }
137 };
138 Trie ac;
139 char buf[1010];
140
141 int main()
142 {
143 // freopen("in.txt","r",stdin);
144 // freopen("out.txt","w",stdout);
145 int n,m;
146 while(scanf("%d%d",&n,&m) == 2)
147 {
148 if(n == 0 && m == 0)break;
149 ac.init();
150 for(int i = 0;i < n;i++)
151 {
152 scanf("%s",buf);
153 ac.insert(buf,1<<i);
154 }
155 for(int i = 0;i < m;i++)
156 {
157 scanf("%s",buf);
158 ac.insert(buf,-1);
159 }
160 ac.build();
161 printf("%d\n",ac.solve(n));
162 }
163 return 0;
164 }View Code
14、ZOJ 3494 BCD Code
这道题很神，数位DP和AC自动机结合，太强大了。
题解here
1 //============================================================================
2 // Name : ZOJ.cpp
3 // Author :
4 // Version :
5 // Copyright : Your copyright notice
6 // Description : Hello World in C++, Ansi-style
7 //============================================================================
8
9 #include <iostream>
10 #include <stdio.h>
11 #include <string.h>
12 #include <algorithm>
13 #include <queue>
14 using namespace std;
15
16 struct Trie
17 {
18 int next[2010][2],fail[2010];
19 bool end[2010];
20 int root,L;
21 int newnode()
22 {
23 for(int i = 0;i < 2;i++)
24 next[L][i] = -1;
25 end[L++] = false;
26 return L-1;
27 }
28 void init()
29 {
30 L = 0;
31 root = newnode();
32 }
33 void insert(char buf[])
34 {
35 int len = strlen(buf);
36 int now = root;
37 for(int i = 0;i < len ;i++)
38 {
39 if(next[now][buf[i]-'0'] == -1)
40 next[now][buf[i]-'0'] = newnode();
41 now = next[now][buf[i]-'0'];
42 }
43 end[now] = true;
44 }
45 void build()
46 {
47 queue<int>Q;
48 fail[root] = root;
49 for(int i = 0;i < 2;i++)
50 if(next[root][i] == -1)
51 next[root][i] = root;
52 else
53 {
54 fail[next[root][i]] = root;
55 Q.push(next[root][i]);
56 }
57 while(!Q.empty())
58 {
59 int now = Q.front();
60 Q.pop();
61 if(end[fail[now]])end[now] = true;
62 for(int i = 0;i < 2;i++)
63 if(next[now][i] == -1)
64 next[now][i] = next[fail[now]][i];
65 else
66 {
67 fail[next[now][i]] = next[fail[now]][i];
68 Q.push(next[now][i]);
69 }
70 }
71 }
72 };
73 Trie ac;
74
75 int bcd[2010][10];
76 int change(int pre,int num)
77 {
78 if(ac.end[pre])return -1;
79 int cur = pre;
80 for(int i = 3;i >= 0;i--)
81 {
82 if(ac.end[ac.next[cur][(num>>i)&1]])return -1;
83 cur = ac.next[cur][(num>>i)&1];
84 }
85 return cur;
86 }
87 void pre_init()
88 {
89 for(int i = 0;i <ac.L;i++)
90 for(int j = 0;j <10;j++)
91 bcd[i][j] = change(i,j);
92 }
93 const int MOD = 1000000009;
94 long long dp[210][2010];
95 int bit[210];
96
97 long long dfs(int pos,int s,bool flag,bool z)
98 {
99 if(pos == -1)return 1;
100 if(!flag && dp[pos][s]!=-1)return dp[pos][s];
101 long long ans = 0;
102 if(z)
103 {
104 ans += dfs(pos-1,s,flag && bit[pos]==0,true);
105 ans %= MOD;
106 }
107 else
108 {
109 if(bcd[s][0]!=-1)ans += dfs(pos-1,bcd[s][0],flag && bit[pos]==0,false);
110 ans %= MOD;
111 }
112 int end = flag?bit[pos]:9;
113 for(int i = 1;i<=end;i++)
114 {
115 if(bcd[s][i]!=-1)
116 {
117 ans += dfs(pos-1,bcd[s][i],flag&&i==end,false);
118 ans %=MOD;
119 }
120 }
121 if(!flag && !z)dp[pos][s] = ans;
122 return ans;
123 }
124
125 long long calc(char s[])
126 {
127 int len = strlen(s);
128 for(int i = 0;i < len;i++)
129 bit[i] = s[len-1-i]-'0';
130 return dfs(len-1,0,1,1);
131 }
132 char str[210];
133 int main()
134 {
135 // freopen("in.txt","r",stdin);
136 // freopen("out.txt","w",stdout);
137 int T;
138 scanf("%d",&T);
139 int n;
140 while(T--)
141 {
142 ac.init();
143 scanf("%d",&n);
144 for(int i = 0;i < n;i++)
145 {
146 scanf("%s",str);
147 ac.insert(str);
148 }
149 ac.build();
150 pre_init();
151 memset(dp,-1,sizeof(dp));
152 int ans = 0;
153 scanf("%s",str);
154 int len = strlen(str);
155 for(int i = len -1;i >=0;i--)
156 {
157 if(str[i]>'0')
158 {
159 str[i]--;
160 break;
161 }
162 else str[i] = '9';
163 }
164 ans -= calc(str);
165 ans %=MOD;
166 scanf("%s",str);
167 ans += calc(str);
168 ans %=MOD;
169 if(ans < 0)ans += MOD;
170 printf("%d\n",ans);
171 }
172 return 0;
173 }View Code
转载于:https://www.cnblogs.com/13224ACMer/p/4873257.html
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