zoukankan      html  css  js  c++  java
  • (简单) POJ 2492 A Bug's Life,二分染色。

    Description

    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.

      判断奇环,二分染色bfs,并查集都可做。

    代码如下:

    // ━━━━━━神兽出没━━━━━━
    //      ┏┓       ┏┓
    //     ┏┛┻━━━━━━━┛┻┓
    //     ┃           ┃
    //     ┃     ━     ┃
    //     ████━████   ┃
    //     ┃           ┃
    //     ┃    ┻      ┃
    //     ┃           ┃
    //     ┗━┓       ┏━┛
    //       ┃       ┃
    //       ┃       ┃
    //       ┃       ┗━━━┓
    //       ┃           ┣┓
    //       ┃           ┏┛
    //       ┗┓┓┏━━━━━┳┓┏┛
    //        ┃┫┫     ┃┫┫
    //        ┗┻┛     ┗┻┛
    //
    // ━━━━━━感觉萌萌哒━━━━━━
    
    // Author        : WhyWhy
    // Created Time  : 2015年07月17日 星期五 20时49分42秒
    // File Name     : 2492.cpp
    
    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <time.h>
    
    using namespace std;
    
    const int MaxN=2010;
    
    int map1[MaxN][MaxN];
    int flag=0;
    int vis[MaxN];
    int N;
    int que[MaxN],first,last;
    
    bool bfs(int u)
    {
        int v;
    
        first=last=0;
        que[last++]=u;
        vis[u]=1;
    
        while(last-first)
        {
            u=que[first++];
    
            for(v=1;v<=N;++v)
                if(map1[u][v]==flag && v!=u)
                {
                    if(vis[v]==vis[u])
                        return 1;
    
                    if(!vis[v])
                    {
                        vis[v]=-vis[u];
                        que[last++]=v;
                    }
                }
        }
    
        return 0;
    }
    
    bool getans()
    {
        memset(vis,0,sizeof(vis));
        first=last=0;
    
        for(int i=1;i<=N;++i)
            if(!vis[i])
                if(bfs(i))
                    return 1;
    
        return 0;
    }
    
    int main()
    {
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
    
        int T,cas=1;
        int Q;
        int a,b;
    
        scanf("%d",&T);
    
        while(T--)
        {
            scanf("%d %d",&N,&Q);
            ++flag;
    
            while(Q--)
            {
                scanf("%d %d",&a,&b);
                map1[a][b]=map1[b][a]=flag;
            }
    
            printf("Scenario #%d:
    ",cas++);
    
            if(getans())
                puts("Suspicious bugs found!
    ");
            else
                puts("No suspicious bugs found!
    ");
        }
        
        return 0;
    }
    View Code
  • 相关阅读:
    Python数据可视化---pygal模块
    Linux常用命令---常用的用户,解压,网络,关机命令
    Python实战---制作专属有声小说(调用百度语音合成接口)
    Linux基本操作---文件搜索命令
    MySQL必知必会1-20章读书笔记
    这是反馈的地方呀
    设计模式--建造者模式
    python 弹窗提示警告框MessageBox
    算法分析设计--递归算法
    Web程序开发最基本的编程模式--MVC编程模式
  • 原文地址:https://www.cnblogs.com/whywhy/p/4655688.html
Copyright © 2011-2022 走看看