poj 3660 Cow Contest(Floyd的拓展应用）

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8968		Accepted: 5033

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

USACO 2008 January Silver

题意;给出m对关系A-B,A能击败B,判断有几只牛能确定排名，

Floyd算法的拓展应用，叫做什么关系闭包，离散数学中的知识，第一次听说

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int map[110][110],n,m,i,j,k;
void floyd()
{
for(k=1;k<=n;k++)
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
if(map[i][k]+map[k][j]>1)//如果i能到达k，k能到达j，则i能 到达j
map[i][j]=1;
}
}
int main()
{
int a,b;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(map,0,sizeof(map));//先把数组清零，表示相互之间不能到达
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
map[a][b]=1;//a能到达b
}
floyd();
int ans=0;
for(i=1;i<=n;i++)
{
int help=0;
for(j=1;j<=n;j++)
if(map[i][j]==1||map[j][i]==1)//判断该点到达的点和被到达的点，即该牛排名之前的牛数量和排名之后的数量
help++;
if(help==n-1)//如果该牛排名之前的牛数量和排名之后的数量和为牛总数减去一，则该牛的排名可以确定
ans++;
}

printf("%d\n",ans);
}
}