zoukankan      html  css  js  c++  java
  • J

    Background 
    Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs. 
    Problem 
    Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.

    Input

    The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.

    Output

    The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.

    Sample Input

    2
    3 3
    1 2
    2 3
    1 3
    4 2
    1 2
    3 4

    Sample Output

    Scenario #1:
    Suspicious bugs found!
    
    Scenario #2:
    No suspicious bugs found!

    Hint

    Huge input,scanf is recommended.
     
     
     
     
    数据量大要用scanf
    #include<iostream>
    #include<map>
    #include<string>
    #include<algorithm>
    #include<cstdio>
    #include<queue>
    #include<cstring>
    #include<cmath>
    #include<vector>
    #include<set>
    #include<queue>
    #include<iomanip>
    #include<iostream>
    using namespace std;
    #define MAXN 2003
    #define INF 0x3f3f3f3f
    typedef long long LL;
    /*带权并查集,
    虫子只和不同性别交配,所以并查集距离为1,如果有不匹配,就标记
    */
    int pre[MAXN],rank[MAXN],n,m;
    int find(int x)
    {
        if(pre[x]==-1)
            return x;
        int fx = pre[x];
        pre[x] = find(pre[x]);
        rank[x] = (rank[x]+rank[fx])%2;
        return pre[x];
    }
    bool mix(int x,int y)
    {
        int fx = find(x),fy=find(y);
        if(fx==fy)
        {
            if(rank[x]!=(rank[y]+1)%2)
                return false;
            return true;
        }
        pre[fy] = fx;
        rank[fy] = (rank[x]-rank[y]+1+2)%2;
        return true;
    }
    void Init()
    {
        memset(rank,0,sizeof(rank));
        memset(pre,-1,sizeof(pre));
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        for(int i=1;i<=t;i++)
        {
            scanf("%d%d",&n,&m);
            Init();
            int x,y;
            bool f = false;
            for(int j=0;j<m;j++)
            {
                scanf("%d%d",&x,&y);
                if(f) continue;
                if(!mix(x,y))
                    f = true;
            }
            printf("Scenario #%d:
    ",i);
            if(f)
                printf("Suspicious bugs found!
    ");
            else
                printf("No suspicious bugs found!
    ");
            printf("
    ");
        }
        return 0;
    }
  • 相关阅读:
    MySQL中的InnoDB中产生的死锁深究
    MySQL中的触发器应用
    你除了在客户端上会使用Cookie,还能使用哪些可以作为数据缓存呢?
    js中实现输入框类似百度搜索的智能提示效果
    linux系统中启动mysql方式已经客户端如和连接mysql服务器
    linux系统安装mysql数据库
    Linux中实用的命令
    Linux下安装jdk中遇到的坑
    Git初始化配置以及配置github
    springboot中配置文件使用2
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6603345.html
Copyright © 2011-2022 走看看