Cow Contest POJ - 3660（floyed求传递闭包）

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
思路：很少写floyed，但是这个算法解决这种所有顶点之间的关系很容易理解。在数据量不大的情况下是可以接受的。
该题思路：
①对于a胜于b，我们使得mp[a][b]=1.
②floyed更新各个点之间的关系。
③遍历每两对点，如果mp[i][j]==1或者mp[j][i]==1的话（i>j或者j>i的个数），计数加一。最后如果计数总数为n-1的话,就代表它和其他n-1个点的关系都确定，那么它的排名就确定。
代码如下：

#include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxx=1e2+10; int mp[maxx][maxx]; int n,m;inline void floyed() {for(int k=1;k<=n;k++)//循环顺序！！！{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;}}} } inline void solve() {int sum=0,ans=0;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(mp[i][j]||mp[j][i]) sum++;}if(sum==n-1) ans++;sum=0;}cout<<ans<<endl; } int main() {memset(mp,0,sizeof(mp));scanf("%d%d",&n,&m);int x,y;while(m--){scanf("%d%d",&x,&y);mp[x][y]=1;}floyed();solve();return 0; }

努力加油a啊，(^o)/~

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