解題思路：給出n頭牛，和這n頭牛之間的m場比賽結果，問最后能知道多少頭牛的排名。 首先考慮排名怎么想，如果知道一頭牛打敗了a頭牛，以及b頭牛打贏了這頭牛，那么當且僅當a+b+1=n時可以知道排名，即為此時該牛排第b+1名。

即推出當一個點的出度和入度的和等于n-1的時候，該點的排名是可以確定的， 即用傳遞閉包來求兩點的連通性，如果d[i][j]==1,那么表示i,j兩點相連通，度數都分別加1

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Cow Contest

Time Limit: 1000MS	?	Memory Limit: 65536K
Total Submissions: 7262	?	Accepted: 4020

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

　　

轉載于:https://www.cnblogs.com/wuyuewoniu/p/4254751.html

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