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 5Sample Output
2 由于出发点不是单源，所以用floyd算法，二维数组赋值为0，若a[i][j]等于1，则代表i，j关系确定，如果j，k关系确定，则i，k关系间接确定 。 #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int n,m; int d[105][105]; //记录奶牛对阵情况 int Floyd() { for(int k=1;k<=n;k++) for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) d[i][j]=d[i][j]||(d[i][k]&&d[k][j]);//记录i和j的关系（直接或间接）是否确定。 } int main() { while(scanf("%d%d",&n,&m)!=EOF) { int a,b; memset(d,0,sizeof(d)); //清空数组 for(int i=1;i<=m;i++) { scanf("%d%d",&a,&b); //若a牛打败b牛，设为1 d[a][b]=1; } Floyd(); int ans=0; //记录牛的数目 for(int i=1;i<=n;i++) { int sum=0; for(int j=1;j<=n;j++) if(d[i][j]||d[j][i]) sum++; if(sum==n-1)//如果一头牛和其他所有牛的关系确定了的话，它的排名也就确定了。 sum++; } cout<<sum<<endl; } return 0; }