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题目

Cow Contest Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 19773 Accepted: 10979 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

分析

碎碎念qwq

最开始我想到的是用拓扑排序，不过迟迟没有思路，看了后豁然开朗awa弗洛伊德的千种套路

正文

题目让我们寻找哪些牛的排名是可以确定的

通过题中给出的案例分析，我们可以得到这样一个关系图

可以看到，2和5的排名是确定的，而1，4，3这三头牛，因为1和4，3的关系不明确，所以这三头牛的关系不确定。

我们又可以从这个图中知道，确定了排名的牛，与其他牛的关系都是确定的， 1，4，3都能打败2，2能打败5 1，2，3，4都能打败5 但是对于1，你只能知道1能打败2，5，但是你不知道3，4和1的关系 所以，我们可以像存图一样，把这个关系存起来 这是一个有向无环图 之后我们就可以用弗洛伊德来更新所有节点之间的关系，找出与其他节点关系确定的节点，就得出最后结果了

代码

#include 
   
    #include 
    
     #include 
     
      #define mt(x, k) memset(x,k, sizeof(x))using namespace std;int main() {       ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);    int mp[110][110], n, m, a, b;    mt(mp, 0);    cin >> n >> m;    for (int i = 0; i < m; ++i) {           cin >> a >> b;        mp[a][b] = 1;    }    for (int k = 1; k <= n; ++k) {           for (int i = 1; i <= n; ++i) {               for (int j = 1; j <= n; ++j) {                   if (mp[i][k] && mp[k][j])mp[i][j] = 1;            }        }    }    int ans = 0, j;    for (int i = 1; i <= n; ++i) {           for (j = 1; j <= n; ++j) {               if (i == j)continue;            if (mp[i][j] == 0 && mp[j][i] == 0)break;        }        if (j > n)ans++;    }    cout << ans << '\n';    return 0;}
     
    
   

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