POJ 3660 图传递闭包
2016-09-19 22:06
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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
Sample Output
View Code
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 floyd拓扑判断连通性,如果出度+入度=n-1则可判断排名。
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <cmath> #include <stdlib.h> #define N 105 using namespace std; int n,m; int g ; int u,v; void floyd() { for(int k = 1;k<=n;k++) for(int i = 1;i<=n;i++) for(int j = 1;j<=n;j++) g[i][j] |= g[i][k]&&g[k][j]; } int main() { //freopen("caicai.txt","r",stdin); scanf("%d%d",&n,&m); for(int i = 0;i<m;i++) { scanf("%d%d",&u,&v); g[u][v] = 1; } floyd(); int ans = 0; for(int i = 1;i<=n;i++) { int cnt = 0; for(int j = 1;j<=n;j++) { if(i==j) continue; if(g[i][j]||g[j][i]) cnt++; } if(cnt==n-1) ans++; } cout<<ans<<endl; return 0; }
View Code
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