Cow Contest POJ 3660 (Floyed ) (最短路专题）

题目描述

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.

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

题目大意：n头牛有 n 个互不相等的等级，给你 n 头牛之间他们等级高低的关系，问你可以确定多少头牛的等级。

解题思路：一头牛可以确定等级当且仅当这一头牛与其他 n - 1 头牛的关系都确定，我们可以用一幅图来描述这些牛之间的关系，即如果 a 比 b 等级高，即 mp[ a ] [ b ] = 1，反之则为 mp [ a ][ b ] = -1。再通过Floyed 进行有条件的松弛，来得到每头牛之间可以得知的所有关系，最后遍历每一头牛与其他牛之间的关系，如果都存在关系就可以确定这头牛的等级。

注意建图时要添加双向边。

代码：

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define ll long long
#define ull unsigned long long 
#define MOD 998244353 
#define INF 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))  
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;

int mp[105][105];
int n,m;
void Floyed()
{
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                     if(mp[i][k]==1&&mp[j][k]==-1){
                        mp[i][j]=1;
                        mp[j][i]=-1;
                     }else if(mp[j][k]==1&&mp[i][k]==-1){
                        mp[i][j]=-1;
                        mp[j][i]=1;
                     }
        
            }
        }
    }
}
int main()
{
    cin>>n>>m;
    mem(mp,INF);
    for(int i=1;i<=n;i++)mp[i][i]=0;
    for(int i=1;i<=m;i++){
        int a,b;
        scanf("%d %d",&a,&b);
        mp[a][b]=1;
        mp[b][a]=-1;
    }
    Floyed();
   /*/ for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cout<<mp[i][j]<<" ";
        }
        cout<<endl;
    } /*/
    int sum=0;
    for(int i=1;i<=n;i++){
        int k=1;
        for(int j=1;j<=n;j++){
           if(mp[i][j]==INF){
              k=0;
              break;
           }
        }
        if(k){
            sum++;
        }
    }
    cout<<sum;
    return 0;
}