题意:给一个元素周期表的元素符号(114种),再给一个串,问这个串能否有这些元素符号组成(全为小写)。

解法1:动态规划

定义:dp[i]表示到 i 这个字符为止,能否有元素周期表里的符号构成。

则有转移方程:dp[i] = (dp[i-1]&&f(i-1,1)) || (dp[i-2]&&f(i-2,2))     f(i,k):表示从i开始填入k个字符,这k个字符在不在元素周期表中。  dp[0] = 1

代码:

//109ms 0KB
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
#define N 50007string single[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"};
string ss[130] = {"he","li","be","ne","na","mg",
"al","si","cl","ar","ca","sc","ti","cr","mn",
"fe","co","ni","cu","zn","ga","ge","as","se",
"br","kr","rb","sr","zr","nb","mo","tc","ru",
"rh","pd","ag","cd","in","sn","sb","te","xe",
"cs","ba","hf","ta","re","os","ir","pt","au",
"hg","tl","pb","bi","po","at","rn","fr","ra",
"rf","db","sg","bh","hs","mt","ds","rg","cn",
"fl","lv","la","ce","pr","nd","pm","sm","eu",
"gd","tb","dy","ho","er","tm","yb","lu","ac",
"th","pa","np","pu","am","cm","bk","cf","es",
"fm","md","no","lr"};int vis[30][30],tag[30];
int dp[N];
char st[N];void init()
{memset(vis,0,sizeof(vis));memset(tag,0,sizeof(tag));for(int i=0;i<14;i++)tag[single[i][0]-'a'] = 1;for(int i=0;i<100;i++)vis[ss[i][0]-'a'][ss[i][1]-'a'] = 1;
}int main()
{int t,len,i;init();scanf("%d",&t);while(t--){scanf("%s",st+1);len = strlen(st+1);memset(dp,0,sizeof(dp));dp[0] = 1;for(i=0;i<len;i++){if(dp[i]){if(tag[st[i+1]-'a'])dp[i+1] = 1;dp[i+2] |= vis[st[i+1]-'a'][st[i+2]-'a'];}}if(dp[len])puts("YES");elseputs("NO");}return 0;
}

View Code

解法2:DFS

搜索时循环的是元素周期表的符号个数。详见代码

代码: (306ms)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
#define N 50007string ss[130] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v","he","li","be","ne","na","mg",
"al","si","cl","ar","ca","sc","ti","cr","mn",
"fe","co","ni","cu","zn","ga","ge","as","se",
"br","kr","rb","sr","zr","nb","mo","tc","ru",
"rh","pd","ag","cd","in","sn","sb","te","xe",
"cs","ba","hf","ta","re","os","ir","pt","au",
"hg","tl","pb","bi","po","at","rn","fr","ra",
"rf","db","sg","bh","hs","mt","ds","rg","cn",
"fl","lv","la","ce","pr","nd","pm","sm","eu",
"gd","tb","dy","ho","er","tm","yb","lu","ac",
"th","pa","np","pu","am","cm","bk","cf","es",
"fm","md","no","lr"};int vis[N];
int len[140];
char st[N];
int Length;
bool Tag;void init()
{int i;for(i=0;i<14;i++)len[i] = 1;for(i=14;i<114;i++)len[i] = 2;
}void dfs(int u)
{if(u == Length)Tag = 1;if(Tag)return;for(int i=0;i<114;i++){int flag = 1;if(u+len[i] <= Length && !vis[u+len[i]]){for(int j=0;j<len[i];j++){if(ss[i][j] != st[u+j]){flag = 0;break;}}if(flag){vis[u+len[i]] = 1;dfs(u+len[i]);}}}
}int main()
{init();int t,i;scanf("%d",&t);while(t--){scanf("%s",st);Length = strlen(st);memset(vis,0,sizeof(vis));Tag = 0;dfs(0);if(Tag)puts("YES");elseputs("NO");}return 0;
}

View Code

解法3:乱搞,模拟。

分成: 单个元素存在与否,与前面匹不匹配,与后面匹不匹配,总共2^3 = 8种情况,然后O(n)扫过去,代码很长。。。

代码:(586ms)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
#define N 50007string single[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"};
string ss[130] = {"he","li","be","ne","na","mg",
"al","si","cl","ar","ca","sc","ti","cr","mn",
"fe","co","ni","cu","zn","ga","ge","as","se",
"br","kr","rb","sr","zr","nb","mo","tc","ru",
"rh","pd","ag","cd","in","sn","sb","te","xe",
"cs","ba","hf","ta","re","os","ir","pt","au",
"hg","tl","pb","bi","po","at","rn","fr","ra",
"rf","db","sg","bh","hs","mt","ds","rg","cn",
"fl","lv","la","ce","pr","nd","pm","sm","eu",
"gd","tb","dy","ho","er","tm","yb","lu","ac",
"th","pa","np","pu","am","cm","bk","cf","es",
"fm","md","no","lr"};char st[N];
int vis[N];int main()
{int t,len,i,j,k;scanf("%d",&t);while(t--){scanf("%s",st);len = strlen(st);int flag = 1;memset(vis,0,sizeof(vis));for(i=0;i<len;i++){if(vis[i])continue;string S = "";S += st[i];for(j=0;j<14;j++){if(single[j] == S)break;}if(j == 14)  //not single
            {if(i > 0 && !vis[i-1]){S = st[i-1]+S;for(j=0;j<100;j++){if(ss[j] == S)break;}if(j != 100)  //pre match
                    {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100)  //back matchvis[i] = 0;else   //back not matchvis[i] = 1;}}else   //pre not match
                    {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100)  //back matchvis[i+1] = 1;else   //back not match
                            {flag = 0;break;}}else{flag = 0;break;}}}else{if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100)  //back matchvis[i+1] = 1;else   //back not match
                        {flag = 0;break;}}else{flag = 0;break;}}}else   //single
            {if(i > 0 && !vis[i-1]){S = st[i-1]+S;for(j=0;j<100;j++){if(ss[j] == S)break;}if(j != 100)  //pre match
                    {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100)  //back matchvis[i] = 0;else   //back not matchvis[i] = 1;}}else   //pre not match
                    {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100)  //back matchvis[i] = 0;else   //back not matchvis[i] = 1;}elsevis[i] = 1;}}else{if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100)  //back matchvis[i] = 0;else   //back not matchvis[i] = 1;}elsevis[i] = 1;}}}if(flag)puts("YES");elseputs("NO");}return 0;
}

View Code

转载于:https://www.cnblogs.com/whatbeg/p/3876636.html

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