CF Round#460D(div2) Substring 题解

题意：

D. Substring

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 to 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.

解法：

这个图是一个DAG。拓扑排序后，用f[x][i]表示第i个字母（从1到26）截止到第x个节点时一共出现了几次。之后把数组里面所有值遍历一遍即可。
DP+DAG的经典组合。

代码：

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N=300010;
char s[N];
int rudu[N];
int n,m;
struct edge
{
int from,to,next;
}e[N];
int front[N];
int cnt=0;
inline void AddEdge(int u,int v)
{
e[++cnt]=(edge){u,v,front[u]},front[u]=cnt;
}
int f[N][26];
queue<int> q;
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
cin>>s[i];
}
for(int i=1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
rudu[v]++;
AddEdge(u,v);
}
for(int i=1;i<=n;i++)
{
if(rudu[i]==0)
{
q.push(i);
f[i][s[i]-'a']=1;
}
}
int rem=n;
while(!q.empty())
{
int qq=q.front();
q.pop();
rem--;
for(int i=front[qq];i;i=e[i].next)
{
int xx=e[i].to;
for(int j=0;j<26;j++)
{
f[xx][j]=max(f[xx][j],f[qq][j]+((s[xx]-'a')==j));
}
rudu[xx]--;
if(!rudu[xx]) q.push(xx);
}
}
if(rem) return puts("-1"),0;
int ans=0;
for(int i=1;i<=n;i++)
{
for(int j=0;j<26;j++)
{
ans=max(ans,f[i][j]);
}
}
printf("%dn",ans);
return 0;
}

1

最后

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