题目描述
You are given a graph with nn nodes and mm 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 33 . Your task is find a path whose value is the largest.
输入输出格式
输入格式：
The first line contains two positive integers n,mn,m ( 1<=n,m<=3000001<=n,m<=300000 ), denoting that the graph has nn nodes and mmdirected edges.
The second line contains a string ss with only lowercase English letters. The ii -th character is the letter assigned to the ii -th node.
Then mm lines follow. Each line contains two integers x,yx,y ( 1<=x,y<=n1<=x,y<=n ), describing a directed edge from xx to yy . Note that xx can be equal to yy and there can be multiple edges between xx and yy . Also the graph can be not connected.
输出格式：
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
输入输出样例
输入样例#1：
5 4 abaca 1 2 1 3 3 4 4 5
输出样例#1：
3
输入样例#2：
6 6 xzyabc 1 2 3 1 2 3 5 4 4 3 6 4
输出样例#2：
-1
输入样例#3：
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
输出样例#3：
4
说明
In the first sample, the path with largest value is 1→3→4→51→3→4→5 . The value is 33 because the letter ‘a‘ appears 33 times.
XJB DP就行了，之前判一下是不是DAG
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#define ll long long
#define maxn 300005
using namespace std;
int ans=0,g[maxn];
int n,m,id[maxn],u,v;
int f[maxn],hd[maxn];
int to[maxn],ne[maxn];
bool vis[maxn];
char s[maxn];
inline int num(char x){
return x-‘a‘;
}
inline bool topsort(){
queue<int> q;
int x,tot=0;
for(int i=1;i<=n;i++) if(!id[i]) q.push(i);
while(!q.empty()){
x=q.front(),q.pop(),tot++;
for(int i=hd[x];i;i=ne[i]) if(!(--id[to[i]]))
q.push(to[i]);
}
return tot==n;
}
int dp(int x){
if(vis[x]) return g[x];
vis[x]=1,g[x]=0;
for(int i=hd[x];i;i=ne[i]) g[x]=max(g[x],dp(to[i]));
g[x]+=f[x];
return g[x];
}
int main(){
scanf("%d%d",&n,&m);
scanf("%s",s+1);
for(int i=1;i<=m;i++){
scanf("%d%d",&u,&v);
to[i]=v,ne[i]=hd[u];
hd[u]=i,id[v]++;
}
if(!topsort()){
puts("-1");
return 0;
}
for(int i=0;i<26;i++){
memset(vis,0,sizeof(vis));
for(int j=1;j<=n;j++) f[j]=(num(s[j])==i);
for(int j=1;j<=n;j++) ans=max(ans,dp(j));
}
printf("%d\n",ans);
return 0;
}
原文地址：https://www.cnblogs.com/JYYHH/p/8474694.html