Legal or Not
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11170    Accepted Submission(s): 5230
Problem Description
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.
 
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.
 
Output
For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".
 
Sample Input
3 2
0 1
1 2
2 2
0 1
1 0
0 0
 
Sample Output
YES
NO

C/C++:

 #include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0xffffff
using namespace std; const int my_max = ;
int n, m, a, b, my_indeg[my_max];
vector <int> my_G[my_max]; void topsort()
{
int my_cnt = ;
while ()
{
int temp = -;
for (int i = ; i < n; ++ i)
if (my_indeg[i] == )
{
temp = i, my_cnt ++, my_indeg[i] = -;
break;
} if (temp == -) break;
for (int i = ; i < my_G[temp].size(); ++ i)
-- my_indeg[my_G[temp][i]];
my_G[temp].clear();
} if (my_cnt == n)
printf("YES\n");
else
{
for (int i = ; i < n; ++ i)
my_G[i].clear();
printf("NO\n");
}
} int main()
{
while (~scanf("%d%d", &n, &m), n || m)
{
memset(my_indeg, , sizeof(my_indeg));
while (m --)
{
scanf("%d%d", &a, &b);
++ my_indeg[a];
my_G[b].push_back(a);
} topsort();
}
return ;
}

hdu 3342 Legal or Not (topsort)的更多相关文章

  1. HDU.3342 Legal or Not (拓扑排序 TopSort)

    HDU.3342 Legal or Not (拓扑排序 TopSort) 题意分析 裸的拓扑排序 根据是否成环来判断是否合法 详解请移步 算法学习 拓扑排序(TopSort) 代码总览 #includ ...

  2. hdu 3342 Legal or Not

    题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3342 Legal or Not Description ACM-DIY is a large QQ g ...

  3. HDU 3342 Legal or Not(判断是否存在环)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3342 Legal or Not Time Limit: 2000/1000 MS (Java/Othe ...

  4. HDU 3342 Legal or Not(拓扑排序判断成环)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3342 题目大意:n个点,m条有向边,让你判断是否有环. 解题思路:裸题,用dfs版的拓扑排序直接套用即 ...

  5. HDU 3342 -- Legal or Not【裸拓扑排序 &amp;&amp;水题 &amp;&amp; 邻接表实现】

    Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tot ...

  6. hdu 3342 Legal or Not(拓扑排序)

    Legal or Not Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total ...

  7. HDU 3342 Legal or Not(有向图判环 拓扑排序)

    Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

  8. HDU 3342 Legal or Not (最短路 拓扑排序?)

    Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

  9. HDU——3342 Legal or Not

    Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tot ...

随机推荐

  1. [BZOJ4990][Usaco2017 Feb]Why Did the Cow Cross the Road II

    Description Farmer John is continuing to ponder the issue of cows crossing the road through his farm ...

  2. 冰释前嫌——转入Android Studio与连接手机无法识别问题

    前言:曾有段时间被AS+gradle虽紧密结合却依然搞不定联网依赖的模样弄的头疼,尝试了各种改代理.改配置均无果,于是坚守Eclipse进行开发学习,结果一方面受制于gradle Android项目的 ...

  3. nginx::基于Nginx+nginx-rtmp-module+ffmpeg搭建rtmp、hls流媒体服务器

    待续 ffmpeg -re -i "/home/bk/hello.mp4" -vcodec libx264 -vprofile baseline -acodec aac -ar 4 ...

  4. HDU 1532 Drainage Ditches(最大流 EK算法)

    题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=1532 思路: 网络流最大流的入门题,直接套模板即可~ 注意坑点是:有重边!!读数据的时候要用“+=”替 ...

  5. ESP8266开发之旅 网络篇⑬ SPIFFS——ESP8266 SPIFFS文件系统

    授人以鱼不如授人以渔,目的不是为了教会你具体项目开发,而是学会学习的能力.希望大家分享给你周边需要的朋友或者同学,说不定大神成长之路有博哥的奠基石... QQ技术互动交流群:ESP8266&3 ...

  6. JVM 知识点补充——永久代和元空间

    之前已经讲过了不少有关 JVM 的内容,今天准备将之前没有细讲的部分进行补充,比如:永久代和元空间. 永久代 Java 的内存中有一块称之为方法区的部分,在 JDK8 之前, Hotspot 虚拟机中 ...

  7. Mysql数据库(三)Mysql表结构管理

    一.MySQL数据类型 1.数字类型 (1)整数数据类型包括TINYINT/BIT/BOOL/SMALLINT/MEDIUMINT/INT/BIGINT (2)浮点数据类型包括FLOAT/DOUBLE ...

  8. django-模板之comment标签(六)

    index.html <!DOCTYPE html> <html lang="en"> <head> <meta charset=&quo ...

  9. Linux wget 批量下载

    需求:已知50个pdf的URL地址,需要批量下载,该怎么办呢? 方案一:使用wget自带的一个功能 -i 选项  从指定文件中读取下载地址,这样的好处是一直是这一个wget进程下载所有pdf,不会来回 ...

  10. NOIP模拟12

    也算是最近几次比较水的一次吧. 考试时看T1像个打表找规律的题,扔了,去看T2,带修莫队??不会,完戏.看了T3,我决定还是去看T1. 看着T1,我突然发现T2是个大水题:主席树就行,不带修,修改时只 ...