九度OJ 1448 拓扑排序问题

题目链接：http://ac.jobdu.com/problem.php?pid=1448

题目描述：

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 20 11 22 20 11 00 0

样例输出：

YESNO

        题目大意是要判断qq群里师徒关系是否合法，师徒关系可以传递，唯一的要求就是不能相互为师徒。这很显然是拓扑排序问题。这里要纪念一下我用C++模板的第三天，以前总是自己写链表队列什么的，细节繁琐而且易出错，现在有了这些模板就省事多了。当然数据结构的基本功绝对是要扎实的这是运用好模板库的条件。本题用的拓扑排序方法是不断寻找入度为零的节点，然后删除这个节点，如果找到的入度为零的节点数等于总结点数则存在拓扑序列，否则没有，也就是不合法。AC代码如下：

#include <stdio.h>#include <vector>#include <queue>using namespace std;vector <int> edge[501];queue <int> Q;int main(){int inDegree[501];int n,m;while(scanf("%d%d",&n,&m)!=EOF){if(n==0 && m==0)break;for(int i=0;i<n;i++){inDegree[i]=0;edge[i].clear();}while(m--){int a,b;scanf("%d%d",&a,&b);inDegree[b]++;edge[a].push_back(b);}while(Q.empty()==false)Q.pop();for(int i=0;i<n;i++){if(inDegree[i]==0)Q.push(i);}int cnt=0;while(Q.empty()==false){int nowp=Q.front();Q.pop();cnt++;for(int i=0;i<edge[nowp].size();i++){inDegree[edge[nowp][i]]--;if(inDegree[edge[nowp][i]]==0){Q.push(edge[nowp][i]);}}}if(cnt==n) puts("YES");else puts("NO");}return 0;}