cf460d2 动态规划+深度优先搜索

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 
121fc
(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

3

input

6 6
xzyabc
1 2
3 1
2 35 44 36 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

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.

/*这道题目是D题，思路是用dfs跑一遍结合动态规划算出一条路径上出现的次数最多的字母（转成数字，比如s[i]-'a'），比较各条路径，特殊情况是

判环，可以用topsort，也可以用dfs设定标志位的方法来判定，详见下dfs1()

#include<bits/stdc++.h>
using namespace std;
const int MAXN=300005;
vector<int> e[MAXN];
char str[MAXN];
int vis[MAXN],cir;
void dfs1(int u)
{
cir|=(vis[u]==1);
if(vis[u])return;
vis[u]=1;
for(int i=0;i<(int)e[u].size();i++)
{
int v=e[u][i];
dfs1(v);
}
vis[u]=2;
}
int dp[MAXN];
void dfs2(int u,int sp)
{
if(vis[u])return;
vis[u]=1;
int mx=0;
for(int i=0;i<(int)e[u].size();i++)
{
int v=e[u][i];
dfs2(v,sp);
mx=max(mx,dp[v]);
}
dp[u]=mx+(str[u]-'a'==sp);
}
int main()
{
int n,m;
scanf("%d%d%s",&n,&m,str+1);
for(int i=1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
e[u].push_back(v);
}
for(int i=1;i<=n;i++)
if(!vis[i])dfs1(i);
if(cir)return 0*printf("-1\n");
int res=0;
for(int _=0;_<26;_++)
{
for(int i=1;i<=n;i++)
dp[i]=vis[i]=0;
for(int i=1;i<=n;i++)
if(!vis[i])dfs2(i,_);
for(int i=1;i<=n;i++)
res=max(res,dp[i]);
}
return 0*printf("%d\n",res);
}