Codeforces 919D：Substring（拓扑排序+DP）

time limit: per test3 seconds
memory limit: per test256 megabytes
inputstandard: input
outputstandard: 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 $n,m (1≤n,m≤300000)$, 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

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

给出有$n$个点和$m$条边的有向图，每个节点上有一个小写字母，求所有通路中出现次数最多的字母出现的次数，如果出现了无数次，输出$-1​$

Solve

用拓扑排序判断图中是否有环，如果有环，那么环上的字母出现的次数为无限多次，输出$-1$。
如果能够进行拓扑排序，对于每次遍历的节点，更新当前路上字母出现次数的最大值
$dp[i][j]$表示到达节点$i$，字母$’a’+j$出现的最大次数

Code

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
#define INF 0x7f7f7f7f
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
int n,m;
char ch[maxn];
int visi[maxn];
// dp[i][j]表示到i位置，字母j出现的最多次数
int dp[maxn][50];
int cnt;
vector<int>v[maxn];
queue<int>que;
void topo()
{
	for(int i=1;i<=n;i++)
		if(!visi[i])
		{
			que.push(i);
			dp[i][ch[i]-'a']++;
		}
	while(!que.empty())
	{
		cnt++;
		int res=que.front();
		que.pop();
		int sz=v[res].size();
		for(int i=0;i<sz;i++)
		{
			visi[v[res][i]]--;
			if(!visi[v[res][i]])
				que.push(v[res][i]);
			for(int j=0;j<26;j++)
				dp[v[res][i]][j]=max(dp[v[res][i]][j],dp[res][j]+(ch[v[res][i]]-'a'==j));
		}
	}
	if(cnt<n)
		cout<<-1<<endl;
	else
	{
		int ans=0;
		for(int i=1;i<=n;i++)
		{
			for(int j=0;j<26;j++)
				ans=max(ans,dp[i][j]);
		}
		cout<<ans<<endl;
	}
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>m;
	cin>>(ch+1);
	ms(visi,0);
	int x,y;
	for(int i=0;i<m;i++)
	{
		cin>>x>>y;
		v[x].push_back(y);
		visi[y]++;
	}
	topo();
	return 0;
}