Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3060    Accepted Submission(s): 1386

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

简单版

#include <stdio.h>
#include <string.h> const int MAX = 100 + 10;
int n,m,G[MAX][MAX],c[MAX];
bool DFS(int u)
{
c[u] = -1;
for(int v = 0;v < n;v++)if(G[u][v])
{
if(c[v]<0) return false;
if(!c[v]&&!DFS(v)) return false;
}
c[u] = 1;
return true;
}
bool TopSort()
{
memset(c,0,sizeof(c));
for(int v = 0;v < n;v++)if(!c[v])
if(!DFS(v)) return false;
return true;
}
int main()
{
while(scanf("%d%d",&n,&m),n)
{
int x,y;
memset(G,0,sizeof(G));
for(int i = 0;i < m;i++)
{
scanf("%d%d",&x,&y);
G[x][y] = 1;
}
if(TopSort())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}

复杂版(邻接表)

#include <iostream>
#include <cstdlib>
#include <queue>
using namespace std; const int MaxVertexNum = 100;
typedef int VertexType;
typedef struct node
{
int adjvex;
node* next;
}EdgeNode;
typedef struct
{
VertexType vertex;
EdgeNode* firstedge;
int count;
}VertexNode;
typedef VertexNode AdjList[MaxVertexNum];
typedef struct
{
AdjList adjlist;
int n,e;
}ALGraph;
bool CreatALGraph(ALGraph *G)
{
int i,j,k;
EdgeNode* s;
cin>>G->n>>G->e;
if(G->n == 0)
return false;
for(i = 0;i < G->n;i++)
{
G->adjlist[i].vertex = i;
G->adjlist[i].firstedge = NULL;
}
for(k = 0;k < G->e;k++)
{
cin>>i>>j;
s = (EdgeNode *)malloc(sizeof(EdgeNode));
s->adjvex = j;
s->next = G->adjlist[i].firstedge;
G->adjlist[i].firstedge = s;
}
return true;
} void TopSort(ALGraph *G)
{
EdgeNode *p;
queue<int> Q;
int i,j,cnt = 0;
for(i = 0;i < G->n;i++)
G->adjlist[i].count = 0;
for(i = 0;i < G->n;i++)
{
p = G->adjlist[i].firstedge;
while(p != NULL)
{
G->adjlist[p->adjvex].count++;
p = p->next;
}
}
for(i = 0;i < G->n;i++)
if(G->adjlist[i].count == 0)
Q.push(i);
while(!Q.empty())
{
i = Q.front();
Q.pop();
cnt++;
p = G->adjlist[i].firstedge;
while(p != NULL)
{
j = p->adjvex;
G->adjlist[j].count--;
if(G->adjlist[j].count == 0)
Q.push(j);
p = p->next;
}
}
if(cnt != G->n)
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
} int main()
{
ALGraph G;
while(CreatALGraph(&G))
TopSort(&G);
return 0;
}

Legal or Not HDU的更多相关文章

  1. Legal or Not HDU - 3342 (拓扑排序)

     注意点: 输入数据中可能有重复,需要进行处理! #include <stdio.h> #include <iostream> #include <cstring> ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. HDU——T 3342 Legal or Not

    http://acm.hdu.edu.cn/showproblem.php?pid=3342 Time Limit: 2000/1000 MS (Java/Others)    Memory Limi ...

  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(拓扑排序)

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

  9. 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 ...

随机推荐

  1. poj 3279 Fliptile(二进制)

    http://poj.org/problem?id=3279 在n*N的矩阵上,0代表白色,1代表黑色,每次选取一个点可以其颜色换过来,即白色变成黑色,黑色变成白色,而且其上下左右的点颜色也要交换,求 ...

  2. hdu 1757 (矩阵快速幂) 一个简单的问题 一个简单的开始

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1757 题意不难理解,当x小于10的时候,数列f(x)=x,当x大于等于10的时候f(x) = a0 * ...

  3. MySQL学习笔记-大纲

    软件程序性能测试在之前<品味性能之道>系列中已经大量提到,讲解了很多测试方法.测试观念.测试思想等等.最近准备深入MySQL进行学习并总结.分别查阅<MySQL性能调优与架构设计&g ...

  4. C++ 的虚析构函数

    当一个基类的指针指向一个派生类的对象,并用该基类的指针去删除或者析构派生类对象时,如果基类的析构函数不是声明为虚函数,那么在析构时基类的析构函数将会被直接调用,派生类的析构函数应为没被调用而导致内存泄 ...

  5. sqli-labs:18-22,http头部注入

    sqli18: uname和passwd被处理了: uagent和ip插入到了数据库: 还带回显. 抓包改包 sqli19: null sqli20: 审计代码,大概如下 当我们正常登录后userna ...

  6. 20155312 2006-2007-2 《Java程序设计》第二周学习总结

    20155312 2006-2007-2 <Java程序设计>第二周学习总结 课堂内容总结 git:版本控制 生活中的容灾备份 归纳思维.实验思维.计算思维 计算机:实现自动化 学会使用快 ...

  7. input.file样式修改

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  8. Quartz(强大的定时器)

    1.关于Quartz的配置文件说明 # # Quartz会优先读取项目下我们自定义这个quartz.properties配置文件 否则会去读取quartzjar包下org.quatrz包# 下面的那个 ...

  9. 【转】Centos7下Yum安装PHP5.5,5.6,7.0

    默认的版本太低了,手动安装有一些麻烦,想采用Yum安装的可以使用下面的方案: 1.检查当前安装的PHP包 yum list installed | grep php 如果有安装的PHP包,先删除他们 ...

  10. PHP中=>是什么意思

    一般用在php数组键名与元素的连接符如:$arr = array('a'=>'123','b'=>'456'); foreach($arr as $key=>$val){//$key ...