找环
来源:互联网 发布:软件体系结构课程 编辑:程序博客网 时间:2024/04/29 07:48
C - Legal or Not
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?
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.
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.
Input
The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
Output
For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".
If it is legal, output "YES", otherwise "NO".
Sample Input
3 20 11 22 20 11 00 0
Sample Output
YESNO
判断环的存在,简单的拓扑排序:
#include <stdio.h>bool a[105][105];int d[105];int n,m;int find(){int i,j,k;for (i=0;i<n;i++){ d[i]=0; for (j=0;j<n;j++) d[i]+=a[j][i];}for (k=0;k<n;k++){ for (i=0;d[i]&&i<n;i++); if (i==n) return 0; for (d[i]=-1,j=0;j<n;j++) d[j]-=a[i][j];}return 1;}int main (){int i,j,k;while (scanf("%d %d",&n,&m)!=EOF){ if (n==0) return 0; for (i=0;i<=n;i++) for (j=0;j<=n;j++) a[i][j]=0; for (k=0;k<m;k++) { scanf("%d %d",&i,&j); a[i][j]=1; } if (find()) printf ("YES\n"); else printf ("NO\n");}return 0;}
这个代码值得研究
0 0
- 找环
- 找环
- dfs找环
- 找环(优化)
- 找环dfs
- 找
- 找
- 找
- Codevs5570 Xor dfs找环
- 快慢指针和找环入口
- [HDU 4324]Triangle LOVE[找环]
- UVa 247 电话圈 floyd找环
- codeforces 711D. Directed Roads 找环
- Codeforces 711D dfs找环
- HDU 1317(spfa找环)
- 有向图找最小的环
- 找因数,找相同
- 找7,找12
- BCM VOIP 线路统计分析
- discuz常用后台函数
- iOS block的用法
- java -D参数简化加入多个jar
- [ACM] hdu 3791 二叉搜索树
- 找环
- pku 1021
- VMWare四种网络模式
- WAS 命令及常见问题
- 面试算法题:乾坤大挪移(链表旋转K个位置)
- 利用metaclass实现python的aop
- windbg调试-句柄泄露
- Linux2.6进程调度分析(1)-调度策略
- windows下如何下载android源码