Codeforces Round #460 (Div. 2)-D. Substring

D. Substring

time limit per test3 seconds

memory limit per test256 megabytes

Problem Description

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.

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解题心得：

1. 比赛的时候读错了题写到崩溃啊。其实题意是每个点用一个字母表示，一个人随机从一个点开始走，获得的值是他走过的路径中遇到的字母（次数最多的那个）的次数。问这个人在途中可能获得的最大值是多少。如果有环输出-1。
2. 写得贼复杂，tarjan判断环，map去除重边，记忆化搜索得到答案。不想说话去角落默默呆着。

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我的智障代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5+100;
vector <int> ve[maxn];
map <pair<int,int>,int> maps;
char s[maxn];
bool find_cir,in[maxn];
int n,m,Max,dfn[maxn],low[maxn],dp[maxn][26];
stack <int> st;

void init(){
    Max = -1;
    scanf("%s",s+1);
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        if(a == b)//自身到自身形成环
            find_cir = true;
        if(maps[make_pair(a,b)] == 233)//去除重边
            continue;
        maps[make_pair(a,b)] = 233;
        ve[a].push_back(b);
    }
}

int tot = 0;
void tarjan(int x){
    dfn[x] = low[x] = ++tot;
    st.push(x);
    in[x] = true;
    for(int i=0;i<ve[x].size();i++){
        int v = ve[x][i];
        if(!dfn[v]){
            tarjan(v);
            low[x] = min(low[x],low[v]);
        }
        else if(in[v])
            low[x] = min(low[x],dfn[v]);
    }
    int num = 0;
    if(low[x] == dfn[x]){
        while(1){
            int now = st.top();
            st.pop();
            in[now] = false;
            num++;
            if(now == x)
                break ;
        }
        if(num > 1){
            find_cir = true;
            return ;
        }
    }
}

void judge_cir(){//用tarjan判断有没有环
    for(int i=1;i<=n;i++)
        if(!dfn[i]){
            tarjan(i);
        }
}

int dfs(int u,int c){
    if(dp[u][c] != -1)
        return dp[u][c];
    int res = 0;
    for(int i=0;i<ve[u].size();i++){
        int v = ve[u][i];
        res = max(res,dfs(v,c));
    }
    res += (s[u]-'a' == c);
    return dp[u][c] = res;
}

void get_Max(){
    memset(dp,-1,sizeof(dp));
    for(int i=1;i<=n;i++)
        for(int j=0;j<26;j++)
            Max = max(Max,dfs(i,j));
    printf("%d",Max);
    return;
}

int main(){
    scanf("%d%d",&n,&m);
    init();
    judge_cir();
    if(find_cir){
        printf("-1");
        return 0;
    }
    get_Max();
}

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轻松过：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5+100;
vector <int> ve[maxn];
int Max = -1,n,m,dp[maxn][30];
char s[maxn];

void init(){
    memset(dp,-1,sizeof(dp));
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        ve[a].push_back(b);
    }
}

int dfs(int u,int c){
    int res = 0;
    if(dp[u][c] == -2){//在之前已经走过这条边，形成环
        printf("-1");
        exit(0);
    }
    if(dp[u][c] != -1)
        return dp[u][c];
    dp[u][c] = -2;
    for(int i=0;i<ve[u].size();i++){
        int v = ve[u][i];
        res = max(res,dfs(v,c));
    }
    res += (s[u]-'a' == c);
    return  dp[u][c] = res;
}

void get_Max(){
    for(int i=1;i<=n;i++)
        for(int j=0;j<26;j++){
            Max = max(Max,dfs(i,j));
        }
    printf("%d",Max);

}

int main(){
    init();
    get_Max();
}