Codeforces 919 D Substring

题目描述

You are given a graph with nn nodes and mm directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is 33 . Your task is find a path whose value is the largest.

输入输出格式

输入格式：

The first line contains two positive integers n,mn,m ( 1<=n,m<=3000001<=n,m<=300000 ), denoting that the graph has nn nodes and mmdirected edges.

The second line contains a string ss with only lowercase English letters. The ii -th character is the letter assigned to the ii -th node.

Then mm lines follow. Each line contains two integers x,yx,y ( 1<=x,y<=n1<=x,y<=n ), describing a directed edge from xx to yy . Note that xx can be equal to yy and there can be multiple edges between xx and yy . Also the graph can be not connected.

输出格式：

Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.

输入输出样例

输入样例#1：

输出样例#1：

输入样例#2：

输出样例#2：

-1

输入样例#3：

输出样例#3：

说明

In the first sample, the path with largest value is 1→3→4→51→3→4→5 . The value is 33 because the letter 'a' appears 33 times.

XJB DP就行了，之前判一下是不是DAG

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#define ll long long
#define maxn 300005
using namespace std;
int ans=0,g[maxn];
int n,m,id[maxn],u,v;
int f[maxn],hd[maxn];
int to[maxn],ne[maxn];
bool vis[maxn];
char s[maxn];inline int num(char x){return x-'a';
}inline bool topsort(){queue<int> q;int x,tot=0;for(int i=1;i<=n;i++) if(!id[i]) q.push(i);while(!q.empty()){x=q.front(),q.pop(),tot++;for(int i=hd[x];i;i=ne[i]) if(!(--id[to[i]]))q.push(to[i]);}return tot==n;
}int dp(int x){if(vis[x]) return g[x];vis[x]=1,g[x]=0;for(int i=hd[x];i;i=ne[i]) g[x]=max(g[x],dp(to[i]));g[x]+=f[x];return g[x];
}int main(){scanf("%d%d",&n,&m);scanf("%s",s+1);for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);to[i]=v,ne[i]=hd[u];hd[u]=i,id[v]++;}if(!topsort()){puts("-1");return 0;}for(int i=0;i<26;i++){memset(vis,0,sizeof(vis));for(int j=1;j<=n;j++) f[j]=(num(s[j])==i);for(int j=1;j<=n;j++) ans=max(ans,dp(j));}printf("%d\n",ans);return 0;}

转载于:https://www.cnblogs.com/JYYHH/p/8474694.html