Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 15540 | Accepted: 8657 |
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
题目意思:有一群牛,厉害的要排在前面,输入a,b表示a比b要厉害,问有多少不能确定排名的牛,一道简单的传递闭包的题。
传递闭包大概就是。。 先将一张图用一个矩阵表示出来,矩阵中的a[i][j]=1表示i~j有一条直接相连的边。
这样就得到一个0/1矩阵。传递闭包算法的目的就是根据以上的初始矩阵,探索出最终的矩阵,
表示根据初始的直接连接关系,从初始矩阵扩展出一个包括间接连接关系的最终矩阵。这个最终矩阵就是传递闭包矩阵
代码如下:
#include <iostream> #include <cstdio> #include <queue> #include <cstring> #include <string> #include <vector> #include <set> #include <algorithm> using namespace std; #define M 20005 #define N 105 int map[N][N]; void fold(int sum) { for(int i=1;i<=sum;i++) for(int j=1;j<=sum;j++) for(int k=1;k<=sum;k++) map[j][k]=map[j][k]||(map[j][i]&&map[i][k]); } int main() { int n,m; cin>>n>>m; int a,b; memset(map,0,sizeof(map)); for(int i=0;i<m;i++) { cin>>a>>b; map[a][b]=1; } fold(n); int res=0; for(int i=1;i<=n;i++) { int ans=0; for(int j=1;j<=n;j++) { if(i==j)continue; if(map[i][j]||map[j][i])ans++; } if(ans==n-1)res++; } cout<<res<<endl; }