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
inputCopy
5 4
abaca
1 2
1 3
3 4
4 5
outputCopy
3
inputCopy
6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4
outputCopy
-1
inputCopy
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
outputCopy
4
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.
给定节点以及边，定义有向路径长度为出现次数最多的字母数量，求最大值；如果出现了环（即任意大的值）输出-1；
拓扑排序处理后进行简单的dp即可，
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<string>
#include<cstring>
#include<set>
#include<queue>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long int ull;
#define eps 1e-5
typedef pair<int,int> pii;
const ll mod=1e9+7;
ll qpow(ll a,ll b)
{
ll ans=1;
while(b){
if(b%2)ans=ans*a%mod;
b=b/2;
a=a*a%mod;
}
return ans;
}
//dp[i][j]表示到点i时，字母j出现的最大次数
int n,m;
char ch[maxn];
int dp[500030][100];
int degr[maxn];
vector<int>Gr[maxn];
int seq[maxn];
int maxx=-inf;
void init()
{
cin>>n>>m>>ch+1;
for(int i=0;i<m;i++){
int x,y;
cin>>x>>y;
degr[y]++;
Gr[x].push_back(y);
}
}
void work()
{
int head=1,tail=0;
for(int i=1;i<=n;i++){
if(degr[i]==0){
seq[++tail]=i;
dp[i][ch[i]-'a']=1;
}
}
while(head<=tail){
int u=seq[head];
for(int i=0;i<Gr[u].size();i++){
int v=Gr[u][i];
degr[v]--;
if(degr[v]==0){
seq[++tail]=v;
}
for(int j=0;j<26;j++){
if(ch[v]-'a'==j){
dp[v][j]=max(dp[v][j],dp[u][j]+1);
}
else dp[v][j]=max(dp[v][j],dp[u][j]);
maxx=max(maxx,dp[v][j]);
}
}
head++;
}
if(tail!=n)cout<<-1<<endl;
else cout<<maxx<<endl;
}
int main()
{
ios::sync_with_stdio(false);
init();
work();
}