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. 利用Veeam保护SAP HANA数据库

    利用Veeam保护SAP HANA数据库 前言 针对越来越多的SAP HANA备份需求,我们Team翻译.整理.借鉴了Veeam 的SAP HANA 大神 Clemens Zerbe 和 Ali Sa ...

  2. Java中package与import

    使用实例: package 一般来说,package语句必须作为源文件的第一条非注释性语句.一个java源文件只能指定一个包,即只能包含一条package语句,该源文件中可以定义多个类,则这些类将全部 ...

  3. Python3+RobotFramework+pycharm环境搭建

    我的环境为 python3.6.5+pycharm 2019.1.3+robotframework3.1.2 1.安装python3.x 略 之后在cmd下执行:pip  install  robot ...

  4. LeetCode初级算法--设计问题02:最小栈

    LeetCode初级算法--设计问题02:最小栈 搜索微信公众号:'AI-ming3526'或者'计算机视觉这件小事' 获取更多算法.机器学习干货 csdn:https://blog.csdn.net ...

  5. .NET中国开发者峰会11.9 下午分会场1 内容解析

    China .NET Conf 2019中国 .NET 开发者峰会即将在上海召开,这次大会是一届完全由社区组织举办的中国.NET 开发者盛会,我们筹备大会之初就定下了大会的主题是“开源.共享.创新”. ...

  6. win7更新,360手机安装谷歌框架

    这两天把11平台被卸载了,不能打竞技场了,很伤心. 成年男子,总要找点有趣的事情去做.我准备洗心革面,好好学习.(巴拉巴拉巴拉一万字.) 首先第一件事情就是重装系统,(由于买了个假显卡,win10以上 ...

  7. 变量 + 数据类型(数字 + 字符串)(day03整理)

    目录 一.上节课回顾 四 编程语言分类 (一) 机器语言 (二)汇编语言 (三) 高级语言 (四) 网络瓶颈效应 五.执行python程序两种方式 (一) 交互式(jupytre) (二) 命令行式( ...

  8. Text 尺寸获取

    获取text在当前文本内容下应该尺寸: 宽度:text.preferredWidth 高度:text.preferredHeight

  9. 第九篇 Flask的before_request和after_request

    Flask我们已经学习很多基础知识了,现在有一个问题 我们现在有一个 Flask 程序其中有3个路由和视图函数,如下: from flask import Flask app = Flask(__na ...

  10. abp中将SqlServer切换为MySQL

    一.移除默认SQL Server相关包 在EntityFrameworkCore项目下移除包Microsoft.EntityFrameworkCore.SqlServer.Microsoft.Enti ...