Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13409 | Accepted: 7465 |
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
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题意:有n头牛比赛,m个比赛结果,最后问你一共有多少头牛的排名被确定了;
其中如果a战胜b,b战胜c,则也可以说a战胜c,即可以传递胜负。求能确定排名的牛的数目。
思路:用Floyd判断连通,失败和胜利场数等于n-1即确定名次
Ac code:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <string>
#include <map>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
#define fi first
#define se second
#define eps 1
const int mod=1000000000+7;
const int maxn=4500+5;
int dis[maxn][maxn];
void Floyd(int n){
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
dis[i][j]=dis[i][j] || (dis[i][k]&&dis[k][j]);
}
}
}
}
int main() {
int n,m;
while(~scanf("%d%d",&n,&m)){
memset(dis,0,sizeof dis);
//for(int i=1;i<=n;i++) dis[i][i]=0;
for(int i=1;i<=m;i++){
int a,b;
scanf("%d%d",&a,&b);
dis[a][b]=1;
}
Floyd(n);
int cnt,ans,res;
cnt=ans=res=0;
for(int i=1;i<=n;i++){
ans=0;
for(int j=1;j<=n;j++){
if(dis[i][j]||dis[j][i]) ans++;
}
if(ans==n-1) res++;
}
printf("%d\n",res );
}
}