hdu 3342 Legal or Not 拓扑排序判断环。 现在的我,除了刷水题,,还能干什么
Legal or Not
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4960 Accepted Submission(s): 2270
We all know a master can have many prentices and a prentice may have a lot of masters too, it‘s legal. Nevertheless,some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian‘s master and, at the same time, 3xian is HH‘s master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.
Please note that the "master and prentice" relation is transitive. It means that if A is B‘s master ans B is C‘s master, then A is C‘s master.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
If it is legal, output "YES", otherwise "NO".
3 2 0 1 1 2 2 2 0 1 1 0 0 0
YES NO
#include <stdio.h> #include <string.h> #define MAX 110 int graph[MAX][MAX] , in[MAX]; bool visited[MAX] ; int main() { int m,n; while(~scanf("%d%d",&n,&m) && n) { memset(graph,0,sizeof(graph)) ; memset(visited,false,sizeof(visited)) ; memset(in,0,sizeof(in)) ; for(int i = 0 ; i < m ; ++i) { int x , y ; scanf("%d%d",&x,&y) ; if(graph[x][y] == 0) { graph[x][y] = 1 ; in[y] ++ ; } } int count = 0 ; for(int i = 0 ;i < n ; ++i) { for(int j = 0 ; j < n ; ++j) { if(!visited[j] && in[j] == 0) { count ++ ; visited[j] = true ; for(int k = 0 ; k < n ; ++k) { if(graph[j][k] == 1) { graph[j][k] = 0 ; in[k]-- ; } } break ; } } } if(count < n) { puts("NO") ; } else { puts("YES") ; } } return 0 ; }
与君共勉
郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。