POJ 3660 Cow Contest 传递闭包+Floyd

时间：2023-03-08 17:46:29
POJ 3660 Cow Contest 传递闭包+Floyd

原题链接：http://poj.org/problem?id=3660

Cow Contest
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8395   Accepted: 4734

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 5

4 3

4 2

3 2

1 2

2 5

Sample Output

2

Source

题意

给你若干牛之间的优劣关系，问你有多少头牛能够被确定排名。

题解

如果有x头牛比当前牛弱，有y头牛比当前牛强，并且x+y=n-1，那么这头牛的排名就被唯一确定了。转化为图论问题，我们若牛a比牛b强，则连接a,b（单向）。运用floyd的思想，令dp[i][j]表示从i能够走到j，即牛i和牛j之间存在强弱关系，那么转移就是dp[i][j]=dp[i][j] | (dp[i][k] & dp[k][j])，跑一发floyd，再统计每个点的度即可。

代码

#include<iostream>

#include<cstring>

#include<vector>

#include<queue>

#include<algorithm>

#define MAX_N 111

using namespace std;

bool d[MAX_N][MAX_N];

int n,m;

void floyd() {

    for (int k = ; k <= n; k++)

        for (int i = ; i <= n; i++)

            for (int j = ; j <= n; j++)

                d[i][j] = d[i][j] | (d[i][k] & d[k][j]);

}

int de[MAX_N];

int main() {

    cin.sync_with_stdio(false);

    cin >> n >> m;

    for (int i = ; i < m; i++) {

        int u, v;

        cin >> u >> v;

        d[u][v] = ;

    }

    floyd();

    int ans = ;

    for (int i = ; i <= n; i++)

        for (int j = ; j <= n; j++)

            de[i] += d[i][j], de[j] += d[i][j];

    for (int i = ; i <= n; i++)if (de[i] == n - )ans++;

    cout << ans << endl;

    return ;

}

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