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:
    洛谷P2770 航空路线问题 最小费用流
    洛谷P2766 最长不下降子序列问题 网络流_DP
    洛谷P2763 试题库问题 最大流
    洛谷P2756 飞行员配对方案问题 网络流_二分图
    洛谷P2045 方格取数加强版 最小费用流
    洛谷P1231 教辅的组成 网络流
    LA3231 Fair Share 二分_网络流
    [CTSC1999]家园 分层图网络流_并查集
    力扣题目汇总(丑数,重复N的元素,求众数)
    力扣题目汇总(最长连续递增序列,旋转图像(中等),宝石与石头)
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6603345.html
Copyright © 2011-2022 走看看