Codeforces 919D-Substring

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概述

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
3

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
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.

题意：给你一个有n个点m条边的有向图，每个点有个小写字母，一条路径的值为该条路径上一个字母出现的最多次数，若为答案为无穷大则输出-1。

解题思路：拓扑排序，同时每个点记录到达该点时每种字母出现的最多的次数，最后进行统计即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>
using namespace std;
#define LL long long
const int INF = 0x3f3f3f3f;
int n, m;
int u, v, cnt, in[300009];
int s[300009], nt[300009], e[300009];
char ch[300009];
int dp[300009][30];
int main()
{
while (~scanf("%d %d", &n, &m))
{
scanf("%s", ch + 1);
memset(s, -1, sizeof s);
memset(in, 0, sizeof in);
memset(dp, 0, sizeof dp);
cnt = 0;
for (int i = 0; i < m; i++)
{
scanf("%d %d", &u, &v);
nt[cnt] = s[u], s[u] = cnt, e[cnt++] = v;
in[v]++;
}
queue<int>q;
for (int i = 1; i <= n; i++)
if (!in[i]) q.push(i), dp[i][ch[i] - 'a']++;
while (!q.empty())
{
int pre = q.front();
q.pop();
for (int i = s[pre]; ~i; i = nt[i])
{
for (int j = 0; j < 26; j++)
dp[e[i]][j] = max(dp[e[i]][j], (j == ch[e[i]] - 'a') + dp[pre][j]);
if (!(--in[e[i]])) q.push(e[i]);
}
}
int flag = 1;
for (int i = 1; i <= n; i++)
if (in[i]) { flag = 0; break; }
if (!flag) { printf("-1n"); continue; }
int ma = 1;
for (int i = 1; i <= n; i++)
for (int j = 0; j < 26; j++)
ma = max(ma, dp[i][j]);
printf("%dn", ma);
}
return 0;
}