POJ3660 Cow Contest（floyd）

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12558 Accepted: 6988

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 54 34 23 21 22 5

Sample Output

2

Source

USACO 2008 January Silver

思路：大致题意为有N只牛，M场比赛的结果，问有多少只牛的排名可以确定。排名可以确定的牛说明它与任何一只牛比较，都是可以知道胜负的，通过题目给出的比赛或者通过前面的比赛结果推出，直接floyd构造传递闭包。

AC代码：

#include <stdio.h>const int N=105;bool map[N][N];void floyd(int n){    for(int k=1;k<=n;k++)    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(map[i][k]&&map[k][j]) map[i][j]=1;            }        }    }}int main(){    int n,m;    scanf("%d%d",&n,&m);    for(int i=0;i<m;i++)    {        int a,b;        scanf("%d%d",&a,&b);        map[a][b]=1;    }    floyd(n);    int ans=n;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            if(i==j)  continue;            if(!map[i][j]&&!map[j][i]) {    ans--;break;}        }    }    printf("%d\n",ans);    return 0;}