Cow Contest【最短路-floyd】

Cow Contest

POJ - 3660

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

Sample Output

题目大意：第一行给出两个数n,m，n代表奶牛的数量，m代表m场比赛，下面m行每行有两个整数a,b，代表本场比赛a能够击败b，（注：若a能击败b，b能击败c，则代表a能击败c）问最后能确定多少头奶牛的排名。

解决方法：通过使用floyd算法，将每两头奶牛的胜负关系确定，如果知道一头奶牛与其他所有奶牛的胜负关系，则可推出这头奶牛的排名。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e3 + 5;
const ll mod = 1e9+7;
bool mapp[maxn][maxn];
int num[maxn];
int n,m;
void floyd()
{rep(k,1,n) {rep(i,1,n) {rep(j,1,n) {if(mapp[i][k]==true&&mapp[k][j]==true)mapp[i][j]=true;}}}
}
int main()
{//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);cin>>n>>m;memset(mapp,false,sizeof(mapp));while(m--){int a,b;cin>>a>>b;mapp[a][b]=true;}floyd();int ans=0;rep(i,1,n) {rep(j,1,n) {if(mapp[i][j]==true||mapp[j][i]==true)num[i]++;}if(num[i]==n-1)ans++;}cout<<ans<<endl;return 0;
}

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