Codeforces 919D 拓扑排序+树形dp

You are given agraph with n nodes and m directed edges. One lowercase letter is assigned to each node. We define a path‘svalue as the number of the most frequently occurring letter. For example, ifletters on a path are "abaca", then thevalue of that path is 3. Your task isfind a path whose value is the largest.

Input

The first linecontains two positive integers n, m (1 ≤ n, m ≤ 300 000), denoting thatthe graph has n nodes and m directed edges.

The second linecontains a string s with onlylowercase 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 adirected edge from x to y. Note that x can be equal toy and there can be multiple edges betweenx and y. Also the graph can be not connected.

Output

Output a single line with a singleinteger 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 withlargest value is 1 → 3 → 4 → 5. The value is 3 because the letter ‘a‘ appears 3 times.

 


【题意】

有一个具有n个节点,m条边的有向图,每个点对应一个小写字母,现在给出每个顶点对应的字母以及有向边的连接情况,求经过的某一条路上相同字母出现的最多次数。如果次数无限大,则输出-1.

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 3e5 + 5;
 
int n ,m;
char a[maxn];  
int ind[maxn];  //入度
int dp[maxn][30]; 
//dp[i][j]表示到i节点的路径中j的最大出现次数
vector <int> v[maxn];  //建树
bool toop()   
{
	queue <int> q;
	for(int i = 1; i <= n; i++)
	{
		if(ind[i] == 0)
		{
			q.push(i);
			dp[i][a[i] - ‘a‘] ++;    //每个点初始状态
			ind[i] = -1;
		}
	}
	int f = 0;
	while(!q.empty())
	{
		int x = q.front();   //父节点
		f++;             //记录树上点的数量
		q.pop();
		for(int i = 0;i < v[x].size();i++)
		{
			int y = v[x][i];   //子节点
			ind[y]--;
			if(ind[y] == 0) 
			{
				q.push(y);
				ind[y] = -1;
			}
			for(int j = 0;j < 26;j++)    
			{
				if(a[y] - ‘a‘ == j)      
					dp[y][j] = max(dp[y][j], dp[x][j] + 1);    
				else
					dp[y][j] = max(dp[y][j], dp[x][j]);   
			}
		}
	}
	return f==n;
}
 
int main()
{
	scanf("%d%d",&n, &m);
	scanf("%s",a+1);
	for(int i = 1;i <= n;i++) v[i].clear();
	memset(ind, 0, sizeof(ind));
	for(int i = 1;i<= m;i++)
	{
		int x, y;
		scanf("%d%d",&x, &y);
		v[x].push_back(y);  
		ind[y]++;
	}
	if(toop() == 0)		printf("-1\n"); 
	else{
		int ans = 0;
		for(int i = 1; i <= n;i ++){
			for(int j = 0; j < 26;j ++){
				//cout<<dp[i][j]<<" ";
				ans = max(ans,dp[i][j]);    
			}//cout<<endl;
		}
		printf("%d\n",ans);
	}
	return 0;

}
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