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

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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