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.

输入:

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.

输出:

For each test case, print in one line the judgement of the messy relationship.If it is legal, output "YES", otherwise "NO".

样例输入:
3 2
0 1
1 2
2 2
0 1
1 0
0 0
样例输出:
YES
NO
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<stack>
using namespace std; vector<int> edge[];
stack<int> s;
int main(){
int m,n;
int in[];
while (cin>>n>>m && !(n==&&m==)){ for (int i=;i<n;i++){
in[i]=;
edge[i].clear();
}
int x,y;
while(m--){
cin>>x>>y;
in[y]++;
edge[x].push_back(y);
} while(!s.empty()) s.pop(); for (int i=;i<n;i++){
if (in[i] == )
s.push(i);
} int ans=;
while (!s.empty()){
int temp=s.top();
s.pop();
ans++;
for (int i=;i<edge[temp].size();i++){
in[edge[temp][i]]--;
if (in[edge[temp][i]] == )
s.push(edge[temp][i]);
}
}
if (ans==n)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl; } return ;
}

一个大错误:一直wa,跟答案比对了很久,难道只因为我用了栈他用了队列?

不相信玄学,继续找

发现是判断输入的m和n是不是0那里出错了,之前的好几道题自己都这么写,并没有错,但是这道就错了

不应该是cin>>n>>m && n!=0&&m!=0而是cin>>n>>m && !(n==0&&m==0)

╮(╯▽╰)╭

Legal or Not ,图的拓扑的更多相关文章

  1. HDU 3342 Legal or Not (图是否有环)【拓扑排序】

    <题目链接> 题目大意: 给你 0~n-1 这n个点,然后给出m个关系 ,u,v代表u->v的单向边,问你这m个关系中是否产生冲突. 解题分析: 不难发现,题目就是叫我们判断图中是否 ...

  2. 【bzoj5017】[Snoi2017]炸弹 线段树优化建图+Tarjan+拓扑排序

    题目描述 在一条直线上有 N 个炸弹,每个炸弹的坐标是 Xi,爆炸半径是 Ri,当一个炸弹爆炸时,如果另一个炸弹所在位置 Xj 满足:  Xi−Ri≤Xj≤Xi+Ri,那么,该炸弹也会被引爆.  现在 ...

  3. loj#2255. 「SNOI2017」炸弹 线段树优化建图,拓扑,缩点

    loj#2255. 「SNOI2017」炸弹 线段树优化建图,拓扑,缩点 链接 loj 思路 用交错关系建出图来,发现可以直接缩点,拓扑统计. 完了吗,不,瓶颈在于边数太多了,线段树优化建图. 细节 ...

  4. 题目1448:Legal or Not(有向无环图判断——拓扑排序问题)

    题目链接:http://ac.jobdu.com/problem.php?pid=1448 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: ...

  5. Paint the Grid Again (隐藏建图+优先队列+拓扑排序)

    Leo has a grid with N × N cells. He wants to paint each cell with a specific color (either black or ...

  6. bzoj5017 炸弹 (线段树优化建图+tarjan+拓扑序dp)

    直接建图边数太多,用线段树优化一下 然后缩点,记下来每个点里有多少个炸弹 然后按拓扑序反向dp一下就行了 #include<bits/stdc++.h> #define pa pair&l ...

  7. 图的拓扑排序,AOV,完整实现,C++描述

    body, table{font-family: 微软雅黑; font-size: 13.5pt} table{border-collapse: collapse; border: solid gra ...

  8. Ordering Tasks UVA - 10305 图的拓扑排序

    John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task i ...

  9. C#实现有向无环图(DAG)拓扑排序

    对一个有向无环图(Directed Acyclic Graph简称DAG)G进行拓扑排序,是将G中所有顶点排成一个线性序列,使得图中任意一对顶点u和v,若边(u,v)∈E(G),则u在线性序列中出现在 ...

随机推荐

  1. 《R语言入门与实践》第三章:R 对象

    在这一章,包含的内容有: R 的数据类型 属性 类(特殊的属性) Ruby 的数据结构 R 数据类型 R 可以识别六种类型的数据类型,分别是: double integer character log ...

  2. Python中怎么读写文件

    python中对文件的操作大概分为三步:打开文件.操作文件(读.写.追加写入).关闭文件. 1.无论对文件做哪种操作,操作前首先要保证文件被打开了,即需要一个打开的操作. 例:open(XXX.txt ...

  3. Echart使用笔记

    一. registerTheme 注册主题,用于初始化实例的时候指定. Echart官网主题下载: http://echarts.baidu.com/download-theme.html 最好的办法 ...

  4. mysql学习之check无效的解决及触发器的使用

    SQL的约束种类: 一.非空约束 not null 二.唯一约束 unique 三.主键约束 四.外键约束 五.check约束 该约束可用于列之间检查语义限制的,实际应用过程中非常常用!! 然鹅,My ...

  5. oracle创建删除表空间

    create [undo|temporary] tablespace orcp datafile|tempfile 'E:\orcle\oracleBaseDir\oradata\orcp\orcp. ...

  6. nigix反向代理

    参考: https://www.cnblogs.com/yycc/p/8185748.html

  7. Oracle单机Rman笔记[4]---RMAN联机备份

    备注:RMAN备份(仅支持基于spfile的备份,不支持基于init.ora配置的备份) 练习:开启ARCHIVELOG模式 \为归档的重做日志被指FRA和单独的归档日志目标 SQL>show ...

  8. eclipese的一些卡顿问题

    一,在jsp页面上输入拼音的时候很卡顿怎么办? 二,我们在复制粘贴等用到ctry的时候会很卡顿,尤其是在jsp页面.干掉以下就好了.

  9. java打印系统时间

    public class Time { public static void main(String[] args) { Date t = new Date(); DateFormat ti = ne ...

  10. java-猜数字

    package com.jijy.circle; import java.util.Scanner; import java.util.Random; public class demo5 { pub ...