题目1448：Legal or Not

题目1448：Legal or Not
时间限制：1 秒内存限制：128 兆特殊判题：否提交：3161解决：1480
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 1
1 0
0 0
样例输出：
YES
NO
问题分析：

代码：

#include <iostream>#include <stdio.h>#include <vector>#include <queue>#define MAX 100/* run this program using the console pauser or add your own getch, system("pause") or input loop */using namespace std;vector<int> edge[MAX];  //邻接链表存储图 queue<int> q;   //临时存放入度为0的点 int main(int argc, char** argv) {    int inDegree[MAX] ; //统计每个结点的入度    int n,m;    while(scanf("%d%d",&n,&m)!=EOF) {        //1初始化        for(int i=0;i<n;i++) {            //所有结点初始入度为0             inDegree[i] = 0;        }        while(q.empty() == false){            //清空队列             q.pop();        }        //2.输入边          while(m--){            //输入m条边             int a,b;             scanf("%d%d",&a,&b) ;             inDegree[b]++; //b的入度加1                      edge[a].push_back(b);         }        //3.入度为0的结点加入到队列中         for(int i=0;i<n;i++){            if(inDegree[i] == 0){                q.push(i);            }        }        //4. 循环选择入度为0的结点进行删除 直到没有入度为0的点为止        int ans = 0; //记录加入拓扑排序中的结点个数         while(q.empty() == false) {            int tmp = q.front();            q.pop();            ans++;             //删除入度为0的点  即将与其相邻的点组成的边删除 入度减1            for(int i=0;i<edge[tmp].size();i++) {                inDegree[edge[tmp][i]]--;                if(inDegree[edge[tmp][i]] == 0){                    q.push(edge[tmp][i]);                }            }        }        if(ans == n){            printf("YES\n");        }else{            printf("NO\n");        }    }    return 0;}