九度OJ 题目1448：Legal or Not （拓扑排序）

最新推荐文章于 2019-07-19 18:29:45 发布

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最新推荐文章于 2019-07-19 18:29:45 发布

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分类专栏： Algorithm 文章标签：九度 OJ 拓扑排序 1448 判断是否存在环路

本文链接：https://blog.youkuaiyun.com/qq_31820885/article/details/62043175

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ACM-DIY是一个和谐的QQ群，成员间通过交流和帮助形成了一张复杂的‘师傅-徒弟’关系网。题目要求判断这个关系网中是否存在非法的循环关系，即某人既是某人的师傅又是其徒弟。拓扑排序可以用于解决这个问题，合法关系应满足师傅可以有多名徒弟，徒弟也可有多个师傅，但不允许存在A既是B的师傅也是B的徒弟的情况。请根据输入数据，利用拓扑排序算法判断关系是否合法。

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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：

样例输出：

YES
NO

该题利用拓扑排序是限定在一个有向无环图上的性质。

将该题抽象为数学模型即为：判断一个有向图是否存在环路

任何情况下，当需要判断某个图是否属于有向无环图是都要立刻联想到拓扑排序

#include <stdio.h>  
#include<queue>  
#include<vector>  
using namespace std;  
  
vector<int> edge[201];  //只有边属性所以不用再定义结构体，使用int类型代表边  
queue<int> q;           //只是用来放入度为0的节点，与其先进先出的特性没有关系，也可使用stack  
int inDegree[201];      //记录节点的入度数  
  
int main()  
{  
    int n,m;  
    while(scanf("%d%d", &n,&m) != EOF) {  
        if(n == 0 && m == 0) break;  
        //初始化  
        for(int i=0;i<n;i++) {  
            inDegree[i] = 0;  
            edge[i].clear();  
        }  
        //清空队列  
        while(!q.empty()) {  
            q.pop();  
        }  
  
        int a,b;  
        while(m--) {  
            scanf("%d%d", &a,&b) ;  //录入a指向b的有向边  
            edge[a].push_back(b);   //将b放到a的链表中  
            inDegree[b]++;          //b节点的入度加1  
        }  
        //遍历找入度为0的节点加入队列  
        for(int i=0;i<n;i++) {  
            if(inDegree[i] == 0) q.push(i);  
        }  
        int count = 0;  //记录弹出队列的节点个数  
        //从队列中弹出节点进行处理  
        while(!q.empty()) {  
            int c = q.front();  //读队头元素  
            q.pop();            //弹出对头元素  
            count++;  
            //遍历这个入度为0的节点指向的所有节点，  
            for(int i=0;i<edge[c].size();i++) {  
                inDegree[edge[c][i]]--;         //将其指向的节点的入度减1  
                if( inDegree[edge[c][i]] == 0) q.push(edge[c][i]);  //若入度变为0，入队列  
            }  
        }  
        //当所有节点都经过队列弹出后说明，存在拓扑排序，即存在有向无环图  
        if(count == n) puts("YES");  
        else puts("NO");  
  
    }  
  
    return 0;  
}