Leagal or Not (九度 OJ 1448)

Leagal or Not (九度 OJ 1448)

Leagal or Not (九度 OJ 1448)

1.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 2
0 1
1 2
2 2
0 11 0
0 0
样例输出：
YES
NO

2.基本思路

该题所要求解的目标用数学的语言进行描述就是，判断该结点和边构成的图是否为有向无环图。即利用拓扑排序的思想，每次取出图中入度为0的结点，从图中去掉该结点，然后去掉以该结点作为弧尾的边，以此反复，直到不存在入度为0的结点，如果此时图中所有的结点均已拓扑排序完毕，那么该图就是有向无环图，否则图中必定存在环路。该算法的时间复杂度为O(N+E)其中N为结点的个数，E为边的数量。

3.代码实现

#include <iostream>
#include <vector>
#include <queue>#define N 101
using namespace std;vector<int> edge[N];//邻接链表,用于存储图中存在的边
queue<int> Q;//该队列用于存储当前图中入度为0的结点
int inDreege[N];//存储每个结点的入度int main()
{int n,m;int n1,n2;while(scanf("%d%d",&n,&m)!=EOF){if(m==0&&n==0)break;for(int i=0;i<n;i++){edge[i].clear();inDreege[i]=0;}for(int i=0;i<m;i++){//输入m条边scanf("%d%d",&n1,&n2);//n1->n2edge[n1].push_back(n2);inDreege[n2]++;}pty()==false){//清空队列Q.pop();}for(int i=0;i<n;i++){//将入度为0的结点加入到队列中if(inDreege[i]==0)Q.push(i);}int cnt=0;//记录已经拓扑排序好的结点个数pty()==false){int t = Q.front();//取队头元素Q.pop();cnt++;for(int i=0;i<edge[t].size();i++){inDreege[edge[t][i]]--;if(inDreege[edge[t][i]]==0)Q.push(edge[t][i]);}}if(cnt==n)//所有的结点均已经拓扑排序完毕printf("YESn");elseprintf("NOn");}return 0;
}/*
样例输入：
3 2
0 1
1 2
2 2
0 1
1 0
0 0
样例输出：
YES
NO
*/