Cow Contest
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 15929 | Accepted: 8888 |
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
Source
Source File
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 110
int win[MAXN][MAXN];
int degree[MAXN];
int main(){
int n,m;
while(scanf("%d %d",&n,&m)==2){
memset(degree,0,sizeof(degree));
memset(win,0,sizeof(win));
for(int i=1;i<=m;i++){
int u,v;
scanf("%d %d",&u,&v);
win[u][v]=1;
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
for(int k=1;k<=n;k++){
if(win[j][i]&&win[i][k])//表达这种传递思想
win[j][k]=1;
}
}
}
int num=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
degree[i]+=win[i][j];
degree[j]+=win[i][j];
}
}
for(int i=1;i<=n;i++)//如果说i+j==n-1，那么就说明这个关系可以确定
if(degree[i]==n-1) num++;
printf("%d\n",num);
}
}
中心思想：N个选手，如果A比B强,B比C强，则A必比C强。告知若干个强弱关系，问有多少人的排名可以确定
比如说，在输入的数据中，4>3,3>2，那么我们就认为4>2,其实这里面有一种传递关系的思想。
假如说i个人可以打败我，而我可以打败j个人，那么这个关系也就随之确定了，进行num自增操作。