Cow Contest

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?9317	?	Accepted:?5249

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

題意：有n頭牛比賽，m種比賽結果，最后問你一共有多少頭牛的排名被確定了，其中如果a戰勝b，b戰勝c，則也可以說a戰勝c，即可以傳遞勝負。求能確定排名的牛的數目。

解題思路：如果一頭牛被x頭牛打敗，打敗y頭牛，且x+y=n-1，則我們容易知道這頭牛的排名就被確定了，所以我們只要將任何兩頭牛的勝負關系確定了，在遍歷所有牛判斷一下是否滿足x+y=n-1，將滿足這個條件的牛數目加起來就是所求解。

可以利用傳遞閉包，將某頭牛到其它牛的"可達性"求出。

#include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxn = 105; int n,m,map[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++)map[i][j] = map[i][j] || (map[i][k] && map[k][j]); }int main() {int u,v;while(scanf("%d%d",&n,&m)!=EOF){memset(map,0,sizeof(map));for(int i = 1; i <= m; i++){scanf("%d%d",&u,&v);map[u][v] = 1;}floyd();int ans = 0,tmp;for(int i = 1; i <= n; i++){tmp = 0;for(int j = 1; j <= n; j++)if(i != j && (map[i][j] || map[j][i]))tmp++;if(tmp == n-1) ans++;}printf("%d\n",ans);}return 0; }

總結

以上是生活随笔為你收集整理的poj 3660(Floyd求传递闭包)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。