TOJ3650 Legal or Not
Legal or Not

Total Submit: 41 Accepted: 18
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
#include <iostream>
#include <queue>
#include <string>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000; struct info {
int id;
bool operator < (const info& a) const {
return id > a.id;
}
}; priority_queue<info> Q;
int gn, gm;
vector<int> g[maxn];
int du[maxn], num = 0; void init() {
num = 0;
int i;
for(i = 0; i < maxn; i++) {
g[i].clear();
}
memset(du, 0, sizeof(du));
while(!Q.empty()) Q.pop();
} bool top_sort() {
int i;
info tmp;
for(i = 0; i < gn; i++) {
if(!du[i]) {
tmp.id = i;
Q.push(tmp);
}
}
while(!Q.empty()) {
info t = Q.top();
Q.pop();
num++;
int x = t.id;
for(i = 0; i < (int)g[x].size(); i++) {
int t = g[x][i];
du[t]--;
if(!du[t]) {
tmp.id = t;
Q.push(tmp);
}
}
}
if(num==gn) {
return true;
}
return false;
} int main()
{
int i;
int from, to;
while(scanf("%d%d", &gn,&gm) != EOF) {
if(gn ==0 && gm==0) break;
init();
for(i = 0; i < gm; i++) {
scanf("%d%d", &from, &to);
g[from].push_back(to);
du[to]++;
}
bool res = top_sort();
if(res) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
}
}
TOJ3650 Legal or Not的更多相关文章
- iOS之App Store上架被拒Legal - 5.1.5问题
今天在看到App Store 上架过程中,苹果公司反馈的拒绝原因发现了这么一个问题: Legal - 5.1.5 Your app uses background location services ...
- 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 ...
- 苹果开发者账号申请时报错提示错误:Legal Entity Name
he information you entered did not match your profile in the D&B database. Before submitting you ...
- 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 ...
- The only legal comparisons are between two numbers, two strings, or two dates.
The only legal comparisons are between two numbers, two strings, or two dates. Left hand operand is ...
- Legal or Not
Legal or Not Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total ...
- hdoj 3342 Legal or Not【拓扑排序】
Legal or Not Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Tota ...
- 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 ...
- Legal or Not(拓扑排序判环)
http://acm.hdu.edu.cn/showproblem.php?pid=3342 Legal or Not Time Limit: 2000/1000 MS (Java/Others) ...
随机推荐
- PHP 解压zip文件的函数封装
/** * zip文件解压 * * @param $zipFilePath zip文件的路径,可以不加zip文件后缀.如果其他类型的文件伪装成zip解压也会失败 * @param $directory ...
- mssql 容易掉进的坑
1. 重复 使用 into #tabel(不是在开头使用insert into ) 会报错 if 1=1 begin select * into #tabel from product ...
- Linux 权限相关
Linux中,所有文件都有 三种权限:User ,Group,Other 三个文件: /etc/passwd :包括所有系统账号,一般用户身份和root信息 /etc/shadow :保存个人密码 / ...
- [POJ] #1006# Biorhythms : 最小公倍数/同余问题
一. 题目 Biorhythms Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 127263 Accepted: 403 ...
- air for ios
在 Adobe AIR 中为不同屏幕尺寸的多种设备提供支持 使用Flash Builder 4.5进行多平台游戏开发 手机屏幕触控技术与提升AIR在Android上的触控体验 AIR Native E ...
- shell学习目录
1. 了解shell 2. shell 入门基础 3. Shell脚本文件中常用的操作语句
- android系统的文件夹选择器
aFileChooser: https://github.com/iPaulPro/aFileChooser/issues, 这个是最适合android的文件选择器,看看有无可能改成文件夹选择器. f ...
- 由于SSH配置文件的不匹配,导致的Permission denied (publickey)及其解决方法。
读者如要转载,请标明出处和作者名,谢谢.地址01:http://space.itpub.net/25851087地址02:http://www.cnblogs.com/zjrodger/作者名:zjr ...
- [转]undefined reference问题总结
转自http://ticktick.blog.51cto.com/823160/431329 最近在Linux下编程发现一个诡异的现象,就是在链接一个静态库的时候总是报错,类似下面这样的错误: (.t ...
- 为何j2ee变成了javaee?
Sun的版本命名有点乱,Java刚面世时还貌不惊人,直到1.2出现后进步很大,Sun就叫它Java 2了,这个称谓持续到1.4,因此Java的三个平台对应的是J2ME(Java 2 Mobile Ed ...