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 5

4 3

4 2

3 2

1 2

2 5

Sample Output

2

Source

USACO 2008 January Silver

題目大意：給定N個人，然后知道這N個人的兩兩對決的情況，前一個為勝者，問M個對決情況中最多能確定幾個人的名次.

思路：想到了Floyd算法，可是怎么來關聯還是不知所以然，網上查了資料后，才明白用Floy求傳遞閉包，只要該人的關系和其他人的關系確定，那么他的名次就是確定的。

Code

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N=110;
int mp[N][N];

int main()
{
int n,m;
while (cin>>n>>m)
{
int i,j,k;
int u,v;
memset(mp,0,sizeof(mp));
for (i=0;i<m;i++)
{
cin>>u>>v;
mp[u][v]=1;
}
for(k=1;k<=n;k++)
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
if(mp[i][k]==1&&mp[k][j]==1) //A和B有關，B和C有關，則Ac有關，不等號的傳遞性

mp[i][j]=1;
//Floyd();
int ans=0;
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
{
if (j==i) continue;
if (mp[i][j]==0&&mp[j][i]==0) break;
}
if (j>n) ans++;
}
cout<<ans<<endl;
}
}

總結

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