| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16341 | Accepted: 9146 |
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
问你根据这些关系可以推出有几头牛的名次是确定的
关系的确定分两种情况：
比如A和C
1.直接确定 题目直接告诉你了A比C厉害
2.间接确定 题目告诉你A比B厉害，B又比C厉害，那么我们可以推出A比C厉害
要求传递闭包的话我们必须确定可达矩阵
可达矩阵：G[i][j]=1的话，代表i和j之间是可以到达的（直接或者间接到达）
#include<stdio.h> #include<iostream> #include<math.h> #include<string.h> #include<set> #include<map> #include<list> #include<queue> #include<algorithm> using namespace std; typedef long long LL; int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31}; int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31}; int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}}; int getval() { int ret(0); char c; while((c=getchar())==' '||c==' '||c==' '); ret=c-'0'; while((c=getchar())!=' '&&c!=' '&&c!=' ') ret=ret*10+c-'0'; return ret; } #define max_v 105 int G[max_v][max_v]; int n,m; void floyd()//求可达矩阵 { for(int k=1;k<=n;k++) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(i==j) continue; if(G[i][k]==1&&G[k][j]==1) G[i][j]=1; } } } } int main() { int x,y; while(~scanf("%d %d",&n,&m)) { memset(G,0,sizeof(G)); for(int i=1;i<=m;i++) { scanf("%d %d",&x,&y); G[x][y]=1; } floyd(); int sum=0; int cnt=0; for(int i=1;i<=n;i++) { cnt=0; for(int j=1;j<=n;j++) { //printf("%d ",G[i][j]); if(G[i][j]||G[j][i]) cnt++; } // printf(" "); if(cnt==n-1)//存在某牛和其他n-1头牛的关系直接或间接确定，那么该牛名次确定 sum++; } printf("%d ",sum); } return 0; } /* 分析；给你n头牛，m组关系，每组关系形式为A B，代表A比B厉害 问你根据这些关系可以推出有几头牛的名次是确定的 如果一头牛和其他的所有牛的关系都确定了的话，那么该牛的名次也是确定的 关系的确定分两种情况： 比如A和C 1.直接确定 题目直接告诉你了A比C厉害 2.间接确定 题目告诉你A比B厉害，B又比C厉害，那么我们可以推出A比C厉害 所有我们必须考虑间接确定的情况，其实这种间接确定就是一个传递闭包 要求传递闭包的话我们必须确定可达矩阵 可达矩阵：G[i][j]=1的话，代表i和j之间是可以到达的（直接或者间接到达） 可达矩阵可以使用floyd算法确定，虽然Floyd是求最短路的，但是也是可以求传递闭包（可达矩阵的） */