2019 ICPC 徐州网络赛

A.00:45:42(-1) solved by zcz

队友说抄个板子好了，我看了一下板子是扩展中国剩余定理

#include<iostream>
#include<cstdio>
#include<climits>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
#define LL long long
const int maxn=1e5+5;
int n;
LL exgcd(LL a,LL b,LL &x,LL &y){if(!b){x=1,y=0;return a;}LL re=exgcd(b,a%b,x,y),tmp=x;x=y,y=tmp-(a/b)*y;return re;
}
LL m[maxn],a[maxn];      //m为模数集，a为余数集
LL exCRT(){LL M=m[1],A=a[1],t,d,x,y;int i;for(i=2;i<=n;i++){d=exgcd(M,m[i],x,y);//解方程if((a[i]-A)%d)return -1;//无解x*=(a[i]-A)/d,t=m[i]/d,x=(x%t+t)%t;//求xA=M*x+A,M=M/d*m[i],A%=M;//日常膜一膜（划掉）模一模，防止爆
    }A=(A%M+M)%M;return A;
}map<long long ,int> isF;int main()
{long long f1=1,f2=1;while(f1<=1e16){isF[f1]=1;long long t=f1;f1=f1+f2;f2=t;}int i,j;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%lld%lld",&m[i],&a[i]);long long t=exCRT();if(t==-1){cout<<"Tankernb!"<<endl;continue;}if(isF[t]==0){cout<<"Zgxnb!"<<endl;}else{cout<<"Lbnb!"<<endl;}}return 0;
}

B.00:11:42 solved by hl

范围小的话直接并查集，这题范围比较大，所以要离散化(雾)

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 3e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
int fa[maxn];
int Hash[maxn];
struct query{int op,x;
}query[maxn];
int find(int x){if(x == fa[x]) return x;return fa[x] = find(fa[x]);
}
void Union(int a,int b){a = find(a); b = find(b);fa[a] = b;
}
int main(){Sca2(N,M);int cnt = 0;for(int i = 1; i <= M ; i ++){query[i].op = read(); query[i].x = read();Hash[++cnt] = query[i].x;if(query[i].op == 1){Hash[++cnt] = query[i].x + 1;}}sort(Hash + 1,Hash + 1 + cnt);cnt = unique(Hash + 1,Hash + 1 + cnt) - Hash;for(int i = 1; i <= cnt + 1; i ++) fa[i] = i;Hash[cnt + 1] = -1;for(int i = 1; i <= M ; i ++){query[i].x = lower_bound(Hash + 1,Hash + 1 + cnt,query[i].x) - Hash; if(query[i].op == 1){Union(query[i].x,query[i].x + 1);}else{Pri(Hash[find(query[i].x)]);}} return 0;
}

C.00:12:36(-3) solved by zcz

枚举题意，枚举到和出题人心意相通为止

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;int main()
{int w;while(cin>>w){if(w%2==0&&w>2){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}}return 0;
}

D.1:05:12 solved by hl

我就比较好奇，出这种题也有钱拿吗？

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 110;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
string S,T;
int main(){cin >> T;Sca(N);while(N--){cin >> S;if(S.length() < T.length()){if(T.find(S) != S.npos) puts("my child!");else puts("oh, child!");}else if(S.length() > T.length()){if(S.find(T) != S.npos) puts("my teacher!");else puts("senior!");}else{if(S == T) puts("jntm!");else puts("friend!");}}return 0;
}

E.00:38:48 solved by hl

从后往前枚举，树状数组维护下标最大的值，树状数组下标为数的大小

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K,cnt;
int a[maxn],Hash[maxn];
int tree[maxn];
void add(int u,int v){for(;u > 0; u -= u & -u) tree[u] = max(tree[u],v);
}
int getmax(int u){int ans = 0;for(;u <= cnt ; u += u & -u) ans = max(ans,tree[u]);return ans;
}
int ans[maxn];
int main(){Sca2(N,M);cnt = 0;for(int i = 1; i <= N ; i ++){a[i] = Hash[++cnt] = read();Hash[++cnt] = a[i] + M ;}sort(Hash + 1,Hash + 1 + cnt);cnt = unique(Hash + 1,Hash + 1 + cnt) - Hash - 1;for(int i = N ; i >= 1; i --){int t = getmax(lower_bound(Hash + 1,Hash + 1 + cnt,a[i] + M) - Hash);//    cout << "bug" << i << endl;//    cout << lower_bound(Hash + 1,Hash + 1 + cnt,a[i] + M) - Hash << endl;//    cout << t << endl;a[i] = lower_bound(Hash + 1,Hash + 1 + cnt,a[i]) - Hash;if(!t) ans[i] = -1;else ans[i] = t - i - 1;//    cout << ans[i] << endl;
        add(a[i],i);}for(int i = 1; i <= N ; i ++){printf("%d",ans[i]);if(i != N) printf(" ");}return 0;
}

F.03:14:29 solved by hl

本场唯一做的一道非签到题，提前感谢一下出题人

题目数据范围给的很奥妙重重，5000个询问和最多为100个k的情况仿佛在暗示什么

果不其然，一个询问点最多往上数100个祖先，枚举每个祖先作为lca对这个点产生的贡献，将会产生一个数的深度的范围，这个点的贡献是他所有子树中深度不超过某个数产生的，同时减去他的祖先对他的影响，每个询问就会变成至多100个询问。

求一个dfs序，题目就转化为了多次询问一个序列中区间所有大于等于p的数的权值和，把所有询问离线出来，用树状数组解决就好了

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 2e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
LL a[maxn];
struct Edge{int to,next;
}edge[maxn * 2];
int head[maxn],tot;
void init(){for(int i = 0 ; i <= N ; i ++) head[i] = -1;tot = 0;
}
void add(int u,int v){edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;
}
int dep[maxn],p[maxn * 2],cnt,fa[maxn];
PII pos[maxn];
void dfs(int u,int la){pos[u].fi = ++cnt;p[cnt] = u;for(int i = head[u]; ~i; i = edge[i].next){int v = edge[i].to;if(v == la) continue;fa[v] = u;dep[v] = dep[u] + 1;dfs(v,u);}pos[u].se = ++cnt;p[cnt] = u;
}
struct Query{int l,r,id;int k,flag;Query(int l = 0,int r = 0,int id = 0,int k = 0,int flag = 0):l(l),r(r),id(id),k(k),flag(flag){}
}q[maxn];
int Stack[maxn];
bool cmp(Query a,Query b){return a.k < b.k;
}
vector<int>Q[maxn];
LL tree[maxn],ans[maxn];
void add(int u,LL v){for(;u < maxn; u += u & -u) tree[u] +=v ;
}
LL getsum(int u){LL sum = 0;for(;u > 0; u -= u & -u) sum += tree[u];return sum;
}
int main(){Sca(N); init(); cnt = 0;for(int i = 1; i <= N ; i ++) a[i] = read();for(int i = 1; i <= N - 1; i ++){int u = read(),v = read();add(u,v); add(v,u);}dep[1] = 0; fa[1] = -1;dfs(1,-1);Sca(M); int o = 0;for(int i = 1; i <= M ; i ++){int u,w; Sca2(u,w);int j,root;Stack[0] = u;for(j = 0,root = u; j < w && fa[root] != -1;){root = fa[root];Stack[++j] = root;}int Mi = -1;for(int k = j ;k >= 0 ; k --){int v = Stack[k];int Mx = dep[v] + w - (dep[u] - dep[v]);if(Mi >= Mx) continue;if(Mi >= 0) q[++o] = Query(pos[v].fi,pos[v].se,i,Mi,-1);if(Mx >= 0) q[++o] = Query(pos[v].fi,pos[v].se,i,Mx,1);Mi = max(Mi,Mx); }}sort(q + 1,q + 1 + o,cmp);int f = 1;for(int i = 1; i <= N ; i ++){Q[dep[i]].pb(i);}for(int i = 0; f <= o ; i ++){for(int j = 0 ; j < Q[i].size(); j ++){int v = Q[i][j];add(pos[v].fi,a[v]);}while(f <= o && q[f].k == i){ans[q[f].id] += getsum(q[f].l - 1) * -1 * q[f].flag;ans[q[f].id] += getsum(q[f].r) * q[f].flag;f++;}}for(int i = 1; i <= M; i ++){Prl(ans[i]);}return 0;
}

G.00:50:53 solved by hl

回文树，bitset维护一下每个结点字母的数量直接冲就行了

不用bitset,建完树之后dfs统计也行，我寻思bitset可能会快一点，空间复杂度也顶住了

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
const int maxn = 3e5 + 10;
const int maxc = 26;
struct Pal_T{int next[maxn][maxc];int fail[maxn],cnt[maxn],num[maxn],len[maxn],S[maxn];bitset<maxc>P[maxn];int last,n,tot;int newnode(int l){for(int i = 0 ; i < maxc; i ++) next[tot][i] = 0;cnt[tot] = num[tot] = 0;len[tot] = l;P[tot].reset();return tot++;}void init(){tot = 0; newnode(0); //偶数根节点newnode(-1); //奇数根节点last = n = 0;S[n] = -1; fail[0] = 1;}int getfail(int x){while(S[n - len[x] - 1] != S[n]) x = fail[x];return x;}void add(int c){c -= 'a';S[++n] = c;int cur = getfail(last);if(!next[cur][c]){int now = newnode(len[cur] + 2);P[now] = P[cur];P[now][c - 'a'] = 1;fail[now] = next[getfail(fail[cur])][c];next[cur][c] = now;num[now] = num[fail[now]] + 1;}last = next[cur][c];cnt[last]++;}LL count(){LL ans = 0;for(int i = tot - 1; i >= 0; i --){cnt[fail[i]] += cnt[i];ans += 1ll * P[i].count() * cnt[i];} return ans;}
}PT;
char str[maxn];
int main(){scanf("%s",str); PT.init();for(int i = 0;str[i]; i ++) PT.add(str[i]);Prl(PT.count());return 0;
}

I.2:01:19 solved by hl

因为原图是个全排列，可以证明即使将所有有关系的数连边，空间复杂度也是nlogn

这就变成了一个区间查询内部包含多少线段的问题了,简单的二位偏序问题，只要把E题敲完的树状数组复制过来就可以了

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int maxm = 2e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
int a[maxn],id[maxn];
struct Line{int l,r;Line(int l = 0,int r = 0):l(l),r(r){}
}line[maxm];
struct Query{int l,r,id;
}q[maxn];
bool cmp(Query a,Query b){return a.r < b.r;
}
bool cmp2(Line a,Line b){return a.r < b.r;
}
int tree[maxn];
void add(int u,int v){for(;u > 0; u -= u & -u) tree[u] += v;
}
int getsum(int u){int ans = 0;for(;u < N; u += u & -u) ans += tree[u];return ans;
}
int ans[maxn];
int main(){Sca2(N,M);for(int i = 1; i <= N; i ++){a[i] = read(); id[a[i]] = i;}int cnt = 0;for(int i = 1; i <= N ; i ++){for(int j = i + i ; j <= N ; j += i){line[++cnt] = Line(id[i],id[j]);if(line[cnt].l > line[cnt].r) swap(line[cnt].l,line[cnt].r);}}for(int i = 1; i <= M ; i ++){q[i].l = read(); q[i].r = read();q[i].id = i;}sort(q + 1,q + 1 + M,cmp);sort(line + 1,line + 1 + cnt,cmp2);int f = 1;for(int i = 1; i <= M ; i ++){while(f <= cnt && line[f].r <= q[i].r){add(line[f++].l,1);}ans[q[i].id] = getsum(q[i].l);}for(int i = 1; i <= M ; i ++){Pri(ans[i]);}return 0;
}

J.2:11:54 solved by gbs

队友写的

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <map>
#include <set>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>using namespace std;typedef long long LL;
const int maxn = 1e6 + 15;
const int mod = 1e9 + 7;
vector<int> pn[maxn + 15] ;
int th_inv[maxn + 15] ;
int high1[maxn+15];
int quick_mi(LL a, LL k)
{LL ans = 1;while (k){if (k & 1){ans *= a;ans %= mod;}a *= a;a %= mod;k >>= 1;}return ans;
}
int the_hight;
int find_hight(int nowi,int fu)
{int nowa = pn[nowi].size();if (fu != 0){nowa--;}if (nowa == 0){return high1[nowi] = 1;}for (unsigned int i=0; i<pn[nowi].size(); i++){if (pn[nowi][i] == fu){continue;}high1[nowi] = max(high1[nowi],find_hight(pn[nowi][i],nowi)+1);}return high1[nowi];
}
int inva(int a)
{if (th_inv[a] == 0)return th_inv[a] = quick_mi(a,mod-2);return th_inv[a];
}int the_ans(int nowi, int fu,int di)
{int now_ans = 0;if (di + high1[nowi] != the_hight){return 1;}int nowa = pn[nowi].size();if (fu != 0){nowa--;}if (nowa == 0){return 0;}for (unsigned int i=0; i<pn[nowi].size(); i++){if (pn[nowi][i] == fu){continue;}now_ans  = (0LL + now_ans  + the_ans(pn[nowi][i],nowi,di+1))%mod;}//cout<<nowi <<' '<<nowa<<' '<<now_ans <<endl;now_ans  = (1LL * now_ans  *inva(nowa))%mod;now_ans  = quick_mi(now_ans ,nowa);//cout<<"*"<<' '<<nowi<<' '<<now_ans <<endl;return now_ans ;}
int main()
{int n;int a, b;cin >> n;for (int i = 0; i < n-1; i++){scanf("%d%d", &a, &b);pn[a].push_back(b);pn[b].push_back(a);}find_hight(1,0);the_hight = high1[1];//printf("%d\n",(the_ans(1,0,0)+mod)%mod  );printf("%d\n",(1-the_ans(1,0,0)+mod)%mod  );#ifdef VSCodesystem("pause");
#endifreturn 0;
}

K.00:49:48 solved by gbs

队友写的

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <map>
#include <set>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>using namespace std;typedef long long LL;int xxnn[1005];
int yynn[1005];
struct pointa{int xi;int yi;pointa(int x1 =0,int y1 =0){xi =x1;yi = y1;}bool operator <(const pointa & s1)const{if (xi!= s1.xi)return this->xi <s1.xi;return this->yi <s1.yi;}bool operator ==(const pointa & s1)const{return (this->xi ==s1.xi)&&(this->yi ==s1.yi);}};int main()
{int n;cin >>n;map<pointa,int>s1;for (int i=0; i<n; i++){scanf("%d%d",&xxnn[i],&yynn[i]);if (s1[pointa(xxnn[i],yynn[i])] >=1){n--;i--;continue;}else{s1[pointa(xxnn[i],yynn[i])]=1;xxnn[i] *= 2;yynn[i] *= 2;}}int fin_ans = 0;int llx;int lly;map<pointa,int>map1;for (int i=0; i<n; i++){for (int j=i; j<n; j++){llx = (xxnn[i]+xxnn[j])/2;lly = (yynn[i]+yynn[j])/2;if (xxnn[i]==xxnn[j] && yynn[i]==yynn[j]){map1[pointa(llx,lly)]+=1;}else{map1[pointa(llx,lly)]+=2;}fin_ans = max(fin_ans,map1[pointa(llx,lly)]);}}printf("%d\n",n-fin_ans);#ifdef VSCodesystem("pause");
#endifreturn 0;
}

L.3:36:27(-4) solved by zcz,hl

预处理出任意两点间，红点朝上翻滚到另一点红点朝上需要的步数

就是个旅行商问题了

状压dp入门门槛

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1 << 16;
const LL INF = 1e18;
const int mod = 1e9 + 7;
int N,M,K;
LL dis[4005][4005];void init(){dis[0][0]=0;int k=3;int cnt=1;for(int i=1;i<=4000;i++){if(cnt%4==0){dis[i][0]=dis[0][i]=cnt;}else{dis[i][0]=k;dis[0][i]=k;}dis[i][1]=dis[1][i]=dis[i][0]+1;k++;cnt++;}for(int i=2;i<=4000;i++){for(int j=2;j<=4000;j++){dis[i][j]=i+j;}}dis[1][1]=6;dis[1][2]=dis[2][1]=5;dis[1][3]=dis[3][1]=6;dis[2][2]=4;
}
PII node[20];
LL MAP[20][20];
LL dp[20][maxn];
int main(){init();int T = read();while(T--){Sca(N);for(int i = 0; i < N ; i ++){node[i].fi = read();node[i].se = read();}for(int i = 0; i < N ; i ++){for(int j = 0; j < N ; j ++){MAP[i][j] = dis[abs(node[i].fi - node[j].fi)][abs(node[i].se - node[j].se)];}MAP[N][i] = MAP[i][N] = dis[abs(node[i].fi)][abs(node[i].se)];}int ed = (1 << N);for(int i = 0 ; i <= N ; i ++){for(int j = 0 ; j < ed; j ++) dp[i][j] = INF;}dp[N][0] = 0;for(int i = 0 ; i < ed; i ++){for(int j = 0 ; j <= N ; j ++){if(dp[j][i] == INF) continue;for(int k = 0; k < N ; k ++){if(i & (1 << k)) continue;int v = i | (1 << k);dp[k][v] = min(dp[k][v],dp[j][i] + MAP[j][k]);}}}LL ans = INF;for(int i = 0 ; i < N ; i ++) ans = min(ans,dp[i][ed - 1]);Prl(ans);}return 0;
}

M.01:47:31(-4) solved by hl

因为眼睛不好使子序列看成了子串，敲完EXKMP还WA了一发

发现a串只要比b串中的某个位置大，a串后面就都符合，前提是当前位置前面需要找的出子序列和b串位置的前缀一样

树状数组维护前缀最大值，表示a串比这个字符大的时候，前缀最多能匹配多少个数，答案就是max(N - i + 1 + max(0 ~ a[i]))

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
char a[maxn],b[maxn];
int tree[maxn];
void add(int u,int v){for(;u <= 27 ; u += u & -u)tree[u] = max(tree[u],v);
}
int getmax(int u){int ans = -1;for(;u > 0; u -= u & -u)ans = max(ans,tree[u]);return ans;
}
int main(){Sca2(N,M);scanf("%s%s",a,b);int cnt = 0,ans = -1;b[M] = 'a' - 1;for(int i = 0 ; i <= 27; i ++) tree[i] = -1;    add(b[0] - 'a' + 2,0);for(int i = 0 ; i < N; i ++){int v = a[i] - 'a' + 2;int t = getmax(v - 1);if(~t) ans = max(ans,N - i + t);if(cnt < M&& a[i] == b[cnt]){cnt++;add(b[cnt] - 'a' + 2,cnt);}}Pri(ans);return 0;
}

转载于:https://www.cnblogs.com/Hugh-Locke/p/11483449.html

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