D. Substring time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard 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 ≤ 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.
http://codeforces.com/problemset/problem/919/D
题意：
就是让你输入n字母，并且有m个关系都是a指向b的格式的，然后题目说路的价值是指一条路中字母最大的重复次数，问你最大是多少。
解题思路：
一开始的时候是想直接用记忆化搜索的，后来发现要判断是否成环，这里两种判环的写法都可以。下面给出两种代码
第一种是拓扑搜索版的，这个和平时的记忆化搜索可能有点区别，它是有点像广搜就是按入度层级递归上去的。
#include<stdio.h>
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
int n,m,res;
const int Maxn = 300005;
char str[Maxn];
int indegree[Maxn];
int _next[Maxn];
int dp[Maxn];
queue<int>q;
vector<int>head;
vector <int> neigh[Maxn];
bool circle()//拓扑排序中的判环法
{
int rco=0;
for(int i=1;i<=n;i++)
{
if(indegree[i]==0)//找出入度为0的点
{
q.push(i);//判环用的队列
head.push_back(i);//搜索用的数组
rco++;//记录出现的次数
}
}
while(!q.empty())
{
int temp=q.front();
q.pop();
for(int i=0;i<neigh[temp].size();i++)
{
int v=neigh[temp][i];
indegree[v]--;
if(indegree[v]==0)//当产生新的入度为零的点的时候入队，并放进搜索用的数组
{
q.push(v);
head.push_back(v);
rco++;//这里也是记录
}
}
}
if(rco<n)//如果最终入度为零的点少于全部的点那证明成环了，建议画图试试看
{
return true;
}
return false;
}
void init(int v)
{
}
int main()
{
ios::sync_with_stdio(false);
while(cin>>n>>m)
{
res=0;
for(int i=1;i<=n;i++)
{
cin>>str[i];
}
for(int i=1;i<=m;i++)
{
int a,b;
cin>>a>>b;
neigh[a].push_back(b);//把b放进a对应的集合中
indegree[b]++;
}
if(circle())
{
cout<<"-1"<<endl;
continue;
}
for (char let = 'a'; let <= 'z'; let++)//一个一个字母对
{
for (int i = head.size()-1; i >=0; i--)//从后面开始搜索保证从最低层级开始递推
{
int v =head[i];
// cout<<"-----------"<<v<<endl;
int my = str[v] == let;
dp[v] = my;
for (int j = 0; j < neigh[v].size(); j++)
{
dp[v] = max(dp[v], dp[neigh[v][j]] + my);
}
res = max(res, dp[v]);
}
}
printf("%d\n", res);
}
} 第二种是深搜递归型的，这种是一条一条路从最深递归上来的。
#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
#include <sstream>
#include <queue>
#include <iomanip>
#include <algorithm>
#pragma comment(linker, "/STACK:16000000")
using namespace std;
const int Maxn = 300005;
int n, m;
char str[Maxn];
vector <int> neigh[Maxn];
int tk[Maxn];
vector <int> seq;
int dp[Maxn];
int res;
bool findLoop(int v)
{
if (tk[v] == 2) return false;
if (tk[v] == 1) return true;
tk[v] = 1;
for (int i = 0; i < neigh[v].size(); i++)
if (findLoop(neigh[v][i])) return true;
tk[v] = 2;
seq.push_back(v);
return false;
}
int main()
{
scanf("%d %d", &n, &m);
scanf("%s", str + 1);
for (int i = 0; i < m; i++)
{
int a, b; scanf("%d %d", &a, &b);
neigh[a].push_back(b);
}
for (int i = 1; i <= n; i++) if (!tk[i])
{
if (findLoop(i))
{
printf("-1\n"); return 0;
}
}
for (char let = 'a'; let <= 'z'; let++)
{
for (int i = 0; i < seq.size(); i++)
{
int v = seq[i];
int my = str[v] == let;
dp[v] = my;
for (int j = 0; j < neigh[v].size(); j++)
{
dp[v] = max(dp[v], dp[neigh[v][j]] + my);
}
res = max(res, dp[v]);
}
}
printf("%d\n", res);
return 0;
}