zoukankan      html  css  js  c++  java
  • legal or not

    Legal or Not

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 8325    Accepted Submission(s): 3918


    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 2 0 1 1 2 2 2 0 1 1 0 0 0
     
    Sample Output
    YES NO
     
    Author
    QiuQiu@NJFU
     
    Source
     
    Recommend
    lcy   |   We have carefully selected several similar problems for you:  1285 2647 3333 3339 3341 
     
    【思路】拓扑排序判断环
    【code】
    #include<iostream>
    #include<cstdio>
    #include<queue>
    #include<cstring>
    #include<vector>
    using namespace std;
    queue<int>q;
    vector<int>vec[101];
    int n,m,cnt,r[101];
    int main()
    {
        while(scanf("%d%d",&n,&m)&&n&&m)
        {
            cnt=0;
            while(!q.empty())q.pop();
            for(int i=0;i<n;i++)
            vec[i].clear();
            memset(r,0,sizeof(r));
            for(int i=1;i<=m;i++)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                vec[x].push_back(y);
                r[y]++;
            }
            for(int i=0;i<n;i++)
            {
                if(!r[i])
                q.push(i); 
            }
            while(!q.empty())
            {
                int now=q.front();q.pop();
                cnt++;
                for(int i=0;i<vec[now].size();i++)
                {
                    r[vec[now][i]]--;
                    if(!r[vec[now][i]])
                    q.push(vec[now][i]); 
                }
            }
            if(cnt!=n)printf("NO
    ");
            else printf("YES
    ");
        }
        return 0;
     } 
  • 相关阅读:
    正则表达式验证日期(多种日期格式)——转载
    NPOI导出Excel——精简版
    Ubuntu下安装部署.NET Core多版本环境
    搭建 nuget 私用库 BaGet
    ubuntu 监听端口
    搭建npm 私用库 verdaccio
    pgsql 性能测试 pgbench
    docker-compose 创建网络
    安装 docker-compose 启动
    dotnet sdk安装以及环境变量设置 Ubuntu 18.04 LTS
  • 原文地址:https://www.cnblogs.com/zzyh/p/6979196.html
Copyright © 2011-2022 走看看