Codeforces Round #460 (Div. 2) D. Substring BFS拓扑排序dp

Posted 2022-12-02 ProLightsfxjh

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.

Source

Codeforces Round #460 (Div. 2)

My Solution

题意：给出一个可能有环可能不连通的图，找出一个路径其上出现最多的字母出现的次数最大，求这个最大值。

BFS、拓扑排序、dp

这题与以前一个求最长路径的题差不多，这里定义状态dp[i][j]表示从某点开始跑到节点i时路径上出现字母j+'a'的最大次数。

只需要按照拓扑序跑一遍dp即可，

ch[v] != j, dp[v][j] = max(dp[v][j], dp[u][j]);

ch[v] != j, dp[v][j] = max(dp[v][j], dp[u][j] + 1);

最后如果度数不为0的点则有环，ans无穷大，置为-1.

否则 ans = max(ans, dp[i][j]).

时间复杂度 O(26*n)

空间复杂度 O(26*n)

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cstring>
using namespace std;
typedef long long LL;
typedef pair<int, int> ii;
const int MAXN = 3e5 + 8;
const LL MOD = 142857;

string s;
LL ch[MAXN];
vector<int> sons[MAXN];
int deg[MAXN];
LL dp[MAXN][26];
queue<ii> que;



inline void bfs()

    int fa, u, v, sz, i, j;
    while(!que.empty())
        u = que.front().first;
        fa = que.front().second;
        que.pop();
        sz = sons[u].size();
        for(i = 0; i < sz; i++)
            v = sons[u][i];
            if(v == fa) continue;
            deg[v]--;
            for(j = 0; j < 26; j++)
                if(ch[v] != j) dp[v][j] = max(dp[v][j], dp[u][j]);
                else dp[v][j] = max(dp[v][j], dp[u][j] + 1);
            
            if(deg[v] == 0)
                que.push(ii(v, u));
            
        
    


int main()

    #ifdef LOCAL
    freopen("d.txt", "r", stdin);
    //freopen("d.out", "w", stdout);

    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    LL n, m, ans, i, j, u, v, sz;
    memset(deg, 0, sizeof deg);
    memset(dp, 0, sizeof dp);
    cin >> n >> m;
    cin >> s;
    for(i = 1; i <= n; i++)
        ch[i] = s[i-1] - 'a';
    
    for(i = 0; i < m; i++)
        cin >> u >> v;
        sons[u].push_back(v);
    
    for(i = 1; i <= n; i++)
        sz = sons[i].size();
        for(j = 0; j < sz; j++)
            deg[sons[i][j]]++;
        
    
    for(i = 1; i <= n; i++)
        if(deg[i] == 0)
            que.push(ii(i, -1));
            dp[i][ch[i]]++;
        
    
    bfs();
    ans = 0;
    for(i = 1; i <= n; i++)
        if(deg[i])
            ans = -1;
            break;
        
        for(j = 0; j < 26; j++)
            ans = max(ans, dp[i][j]);
        
    

    cout << ans << endl;
    return 0;

  Thank you!

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