Cow Contest(Poj 3660)

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

传递闭包，利用floyd-warshall算法解决

//============================================================================// Name        : test.cpp// Author      : Qihan// Version     :// Copyright   : Your copyright notice// Description : Hello World in C++, Ansi-style//============================================================================#include <stdio.h>#include <stdlib.h>#include <iostream>#include <string.h>using namespace std;typedef long long int LLI;typedef pair<LLI,LLI> PII;#define Lowbit(x) (x & (-x))const int inf = 1000000000;const int maxn = (2048 + 10);bool vis[maxn][maxn];int main() {//    freopen("/home/qihan/Documents/in","r",stdin);    int n,m;    memset(vis,false,sizeof(vis));    scanf("%d%d",&n,&m);    for(int i = 1; i <= n; i ++) {        vis[i][i] = true;    }    for(int i = 1; i <= m; i ++) {        int x,y;        scanf("%d%d",&x,&y);        vis[x][y] = true;    }    for(int k = 1; k <= n; k ++) {        for(int i = 1; i <= n; i ++) {            for(int j = 1; j <= n; j ++) {                if(vis[i][k] == true && vis[k][j] == true)                    vis[i][j] = true;            }        }    }    int ans = 0;    for(int i = 1; i <= n; i ++) {        bool flag = true;        for(int j = 1; j <= n; j ++) {            if(i == j)  continue;            else if(!vis[i][j] && !vis[j][i]) {                flag = false;                break;            }        }        if(flag == true)    ans ++;    }    printf("%d\n",ans);    return 0;}