Codeforces Round #460 (Div. 2)D. Substring

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链接：http://codeforces.com/contest/919/problem/D

D. Substring
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given a graph with n nodes and m 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 3. Your task is find a path whose value is the largest.
Input

The first line contains two positive integers n, m (1 ≤ n, m ≤ 300 000), denoting that the graph has n nodes and m directed edges.

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

Then m lines follow. Each line contains two integers x, y (1 ≤ x, y ≤ n), describing a directed edge from x to y. Note that x can be equal to y and there can be multiple edges between x and y. Also the graph can be not connected.
Output

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

5 4
abaca
1 2
1 3
3 4
4 5

Output

Input

6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4

Output

-1

Input

10 14
xzyzyzyzqx
1 2
2 4
3 5
4 5
2 6
6 8
6 5
2 10
3 9
10 9
4 6
1 10
2 8
3 7

Output

Note

In the first sample, the path with largest value is 1 → 3 → 4 → 5. The value is 3 because the letter ‘a’ appears 3 times.
【题意】

有一个具有n个节点，m条边的有向图，每个点对应一个小写字母，现在给出每个顶点对应的字母以及有向边的连接情况，求经过的某一条路上相同字母出现的最多次数。如果次数无限大，则输出-1.

#include<bits/stdc++.h>
using namespace std;
#define N 300010
#define LL long long

int n,m;
vector<int>g[N];
int in[N];
int dp[N][26];
int pre(){
    queue<int>q;
    int cnt=0;
    for(int i=1;i<=n;++i){
        if(in[i]==0)q.push(i),in[i]--;
    }
    while(!q.empty()){
        int u=q.front();q.pop();
        cnt++;
        for(int i=0;i<g[u].size();++i){
            int v=g[u][i];
            in[v]--;
            if(in[v]==0){
                q.push(v);
            }
        }
    }
    if(cnt!=n)return 1;
    return 0;
}
char str[N];
int tmp[N];
int dfs(int u,int cc){
    if(dp[u][cc]!=-1)return dp[u][cc];
    dp[u][cc]=(str[u]-'a'==cc?1:0);
    for(int i=0;i<g[u].size();++i){
        int v=g[u][i];
        dp[u][cc]=max(dp[u][cc],dfs(v,cc)+(str[u]-'a'==cc?1:0));
    }
    return dp[u][cc];
}
int main(){
    //freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    scanf("%s",str+1);
    for(int j=1;j<=m;++j){
        int u,v;scanf("%d%d",&u,&v);
        g[u].push_back(v);in[v]++,tmp[v]++;
    }
    if(pre())puts("-1");
    else{
        memset(dp,-1,sizeof(dp));
        int ans=0;
        for(int i=1;i<=n;++i){
            for(int j=0;j<26;++j){
                ans=max(ans,dfs(i,j));
                cout<<dfs(i,j)<<' ';
            }puts("");
            //ans=max(ans,dfs(i,str[i]-'a'));
        }
        cout<<ans;
    }
}