POJ 3660 Cow Contest （floyd求联通关系）

Cow Contest

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122685#problem/H

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.

</big>


 




##Input
<big>
* 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
</big>






##Output
<big>
* Line 1: A single integer representing the number of cows whose ranks can be determined
</big>
 
 
 
##Sample Input
<big>
5 5
4 3
4 2
3 2
1 2
2 5
</big>


##Sample Output
<big>
2
</big>

##Hint
<big>
</big>





<br/>
##题意：
<big>
给出N个点，M个点对：
每条点对 A B 意味着A点的权值大于B点.
现在要对这些点进行权值排名，求有多少个点的排名能够确定.
</big>


<br/>
##题解：
<big>
将样例画一遍就比较容易看出来：
若某点跟其他n-1个点都联通，则这个点的排名可以确定. 否则不能.
问题就转换为了求n个点之间的联通关系.
而floyd算法正好可以求任意两点的联通关系，只需要把求最短路时的松弛操作修改一下即可.
dis[i][j] = dis[i][j] || (dis[i][k] && dis[k][j]);
</big>




<br/>
##代码：
``` cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define mid(a,b) ((a+b)>>1)
#define LL long long
#define maxn 110
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n, m;
bool dis[maxn][maxn];

void floyd() {
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                dis[i][j] = dis[i][j] || (dis[i][k] && dis[k][j]);
}

int main(int argc, char const *argv[])
{
    //IN;

    while(scanf("%d %d", &n,&m) != EOF)
    {
        memset(dis, 0, sizeof(dis));
        for(int i=1; i<=n; i++) dis[i][i] = 1;

        for(int i=1; i<=m; i++) {
            int u,v; scanf("%d %d", &u,&v);
            dis[u][v] = 1;
        }

        floyd();

        int ans = 0;
        for(int i=1; i<=n; i++) {
            int cnt1=0, cnt2=0;
            for(int j=1; j<=n; j++) {
                if(i == j) continue;
                if(dis[i][j]) cnt1++;
                if(dis[j][i]) cnt2++;
            }
            if(cnt1+cnt2 == n-1) ans++;
        }

        printf("%d
", ans);
    }

    return 0;
}