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.
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 xand y. Also the graph can be not connected.
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
5 4 abaca 1 2 1 3 3 4 4 5
3
6 6 xzyabc 1 2 3 1 2 3 5 4 4 3 6 4
-1
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
4
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：n个点，m条边，每个点对应字符串位置的字母，一条路径的权值等于这条路径上最大相同字母的个数，求路径的最大权（当权值不固定，即出现环的时候输出-1）
可以把整个图看做一颗树，然后每个节点存储由它为起点的路径中每个字母出现的最大次数，回溯刷新最大值即可
dp[i][j]表示以i为起点的所有路径中字母j+'a'最大出现了多少次
#include<bits/stdc++.h>
using namespace std;
#define Accel ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
const int N=3e5+500;
vector<int>g[N];
string s;
int vis[N],dp[N][26];
void dfs(int u)
{
vis[u]=1;
dp[u][s[u-1]-'a']=1;
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
if(vis[v]==1)
{
cout<<"-1";
exit(0);
}
else
{
if(!vis[v])dfs(v);
for(int k=0;k<26;k++)
dp[u][k]=max(dp[u][k],dp[v][k]+(s[u-1]-'a'==k));
}
}
vis[u]=2;
}
int u,v,ans,n,m;
int main()
{
Accel;
cin>>n>>m>>s;
for(int i=0;i<m;i++)
{
cin>>u>>v;
g[u].push_back(v);
}
for(int i=1;i<=n;i++)
{
if(!vis[i])
{
dfs(i);
for(int j=0;j<26;j++)
ans=max(ans,dp[i][j]);
}
}
cout<<ans;
return 0;
}