HDU 3342 Legal or Not

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3342

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4592    Accepted Submission(s): 2088

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 20 11 22 20 11 00 0

 

Sample Output

YESNO

 

Author

QiuQiu@NJFU

 

Source

HDOJ Monthly Contest – 2010.03.06 

我英语是个渣，但是看懂题目意思还是卓卓有余的，嘿嘿。

题目大意：

ACM-DIY群里面有很多“神牛”，群里的气氛也很和谐，每次有人提问，都有热心的神牛回答，问题问多了，很自然，就有了师徒关系，例如A是B的师父，B是C的师父，那么我们认为A也是C的师父，这种关系是正常的，如果A是B的师父，B也是A的师父，这种关系是不正常的。现在题目给你一些关系，叫你判断这其中存不存在不正常的关系，有不正常关系的话输出NO,否则输出YES。

拓扑排序大水题。只需排完拓扑排序之后看看有没有点没有排到。没有排到说明到最后其它边都没删光了还存在一些点入度不为0的点。也就是关系不正常喽。

下面小的供上AC代码，给路过的好学的孩子们参考参考~

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<sstream>#include<vector>#include<map>#include<list>#include<set>#include<queue>using namespace std;const int maxn=1005,maxe=100005,inf=1<<29;struct node//前向星（静态链表）{    int to,next,w;}edge[maxe];int n,m,Count;int head[maxn],cnt;int in[maxn],id[maxn],iq;//拓扑排序部分数据定义void add(int from,int to){    edge[cnt].to=to;    edge[cnt].next=head[from];    head[from]=cnt++;}void topology(){    queue<int>q;    for(int i=0;i<n;i++)        if(!in[i]) q.push(i);    while(!q.empty())    {        int t=q.front();q.pop();        Count++;        for(int j=head[t];j!=-1;j=edge[j].next)        {            in[edge[j].to]--;            if(in[edge[j].to]==0) q.push(edge[j].to);        }    }}void init(){    Count=0;cnt=0;    memset(in,0,sizeof(in));    memset(head,-1,sizeof(head));}int main(){    while(~scanf("%d%d",&n,&m)&&(n+m))    {        init();        for(int i=0;i<m;i++)        {            int from,to;            scanf("%d%d",&from,&to);            in[from]++;            add(to,from);        }        topology();        //printf("%d\n",Count);        if(Count<n) printf("NO\n");        else printf("YES\n");    }    return 0;}