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≤300000), 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.
就是要我们找到一条路径,里面出现最多的字母,出现的次数最多。
很明显,如果有环,直接输出-1
如果没有环,直接拓扑排序递推即可。就相当于在DAG图上DP。
AC代码:
#pragma GCC optimize(2)
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=3e5+10;
int n,m,dag[N],val[N][30],cnt,res;
char str[N];
int head[N],nex[N],w[N],to[N],tot;
inline void add(int a,int b){to[++tot]=b; nex[tot]=head[a]; head[a]=tot;}
void top_sort(){
queue<int> q; for(int i=1;i<=n;i++) if(!dag[i]) q.push(i),cnt++;
while(q.size()){
int u=q.front(); q.pop();
for(int i=head[u];i;i=nex[i]){
if(--dag[to[i]]==0) q.push(to[i]),cnt++;
for(int j=0;j<26;j++)
val[to[i]][j]=max((str[to[i]]-'a'==j)+val[u][j],val[to[i]][j]);
}
}
if(cnt!=n){puts("-1"); exit(0);}
}
signed main(){
cin>>n>>m; scanf("%s",str+1);
for(int i=1;i<=n;i++) val[i][str[i]-'a']++;
for(int i=1,a,b;i<=m;i++) cin>>a>>b,add(a,b),dag[b]++;
top_sort();
for(int i=1;i<=n;i++) for(int j=0;j<26;j++) res=max(res,val[i][j]);
cout<<res<<endl;
return 0;
}