POJ 3660 Cow Contest (闭包传递)
2015-05-19 16:55
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Cow Contest
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
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7690 | Accepted: 4288 |
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 刚才学了下闭包传递,简单来说就是用Floyd来求两点是否连通。这题里,如果一点的入度加上出度等于N-1,那么就可确定这点的等级,因为除了它自己之外的所有节点要么在它之前要么在它之后,不管它前面的节点后后面的节点怎么排序,都不会影响到它。
#include <iostream> #include <cstdio> using namespace std; const int SIZE = 105; int N,M; int IN[SIZE],OUT[SIZE]; bool G[SIZE][SIZE]; int main(void) { int from,to; int ans; while(scanf("%d%d",&N,&M) != EOF) { fill(&G[0][0],&G ,false); for(int i = 0;i <= N;i ++) IN[i] = OUT[i] = 0; for(int i = 0;i < M;i ++) { scanf("%d%d",&from,&to); G[from][to] = true; } 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][j] || G[i][k] && G[k][j]; for(int i = 1;i <= N;i ++) for(int j = 1;j <= N;j ++) if(G[i][j]) { IN[j] ++; OUT[i] ++; } ans = 0; for(int i = 1;i <= N;i ++) if(IN[i] + OUT[i] == N - 1) ans ++; printf("%d\n",ans); } return 0; }
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