title: Cow Contest 弗洛伊德+传递闭包 nyoj211
tags: [弗洛伊德,传递闭包]
题目链接
描述
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.
- 输入
- 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 BThere are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20.
- 输出
For every case:* Line 1: A single integer representing the number of cows whose ranks can be determined - 样例输入
5 5
4 3
4 2
3 2
1 2
2 5
0 0 - 样例输出
2
分析：
使用Floyd算法来判断传递闭包。
首先使用Floyd算法来求出每两个点的最短路径
然后对于能否确定一个牛的位置的判断，我们假设现在一共有5个牛，对于牛A，如果有两个牛可以打败它并且它又可以打败其他的两个牛的话，我们就可以确定它的位置在3号。
所以对于N个牛，如果能打败它和它能打败的牛的总个数为N-1的话，那么它的位置就可以确定出来，这就叫做传递闭包（以前完全没有听说过）。
代码：
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
#define INF 0x3f3f3f3f
int N,M;
int Tu[102][102];
void Floyd()///弗洛伊德算法，三重for循环
{
for(int k=1; k<=N; k++)
for(int i=1; i<=N; i++)
for(int j=1; j<=N; j++)
{
if(i!=j&&Tu[i][k]!=INF&&Tu[k][j]!=INF&&Tu[i][j]>Tu[i][k]+Tu[k][j])
Tu[i][j]=Tu[i][k]+Tu[k][j];
}
}
int main()
{
int a,b;
while(~scanf("%d%d",&N,&M),N,M)
{
memset(Tu,INF,sizeof(Tu));
while(M--)
{
scanf("%d%d",&a,&b);
Tu[a][b]=1;///表示的是a能战胜b
}
Floyd();
int ans=0;
int sum=0;
for(int i=1; i<=N; i++)
{
sum=0;
for(int j=1; j<=N; j++)
{
if(i!=j)
{
if(Tu[i][j]!=INF)///i能战胜j
sum++;
if(Tu[j][i]!=INF)///j能战胜i
sum++;
}
}
if(sum==N-1)
ans++;
}
printf("%d
",ans);
}
return 0;
}