poj 3660 Cow Contest
2016-03-06 16:27
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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
可以说得上是最短路吧,反正我会是用弗洛伊德暴力水过的,代码很简单,思路清晰就好。
题意:牛之间有绝对的强弱,给出一些胜负关系,问有多少头牛可以确定其绝对排名。
附上代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8987 | Accepted: 5046 |
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
可以说得上是最短路吧,反正我会是用弗洛伊德暴力水过的,代码很简单,思路清晰就好。
题意:牛之间有绝对的强弱,给出一些胜负关系,问有多少头牛可以确定其绝对排名。
附上代码:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { int n,m,i,j,k; while(~scanf("%d%d",&n,&m)) { int a,b,map[105][105],t; memset(map,0,sizeof(map)); for(i=1; i<=m; i++) { scanf("%d%d",&a,&b); map[a][b]=1; } for(i=1; i<=n; i++) for(j=1; j<=n; j++) for(k=1; k<=n; k++) if(map[j][i]&&map[i][k]) //暴力弗洛伊德,所有的关系连成图 map[j][k]=1; int ans=0; for(i=1; i<=n; i++) { int t=0; for(j=1; j<=n; j++) t+=map[i][j]+map[j][i]; if(t==n-1) ans++; } printf("%d\n",ans); } return 0; }
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