zoukankan      html  css  js  c++  java
  • poj1703(并查集,维护两个集合)

    Find them, Catch them
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 53671   Accepted: 16449

    Description

    The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)

    Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:

    1. D [a] [b]
    where [a] and [b] are the numbers of two criminals, and they belong to different gangs.

    2. A [a] [b]
    where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.

    Input

    The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.

    Output

    For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs." and "Not sure yet."

    Sample Input

    1
    5 5
    A 1 2
    D 1 2
    A 1 2
    D 2 4
    A 1 4
    

    Sample Output

    Not sure yet.
    In different gangs.
    In the same gang.
    
    分析:这个题跟poj2492差不多。

    #include<cstdio>
    #include<cstring>
    using namespace std;
    int N;
    int pre[200200];
    void init()
    {
        for(int i=0;i<=N*2;i++)
        pre[i]=i;
    }
    int find(int x)
    {
        if(x!=pre[x])
        pre[x]=find(pre[x]);
        return pre[x];
    }
    void unit(int x,int y)
    {
        int rx=find(x),ry=find(y);
        if(rx<ry) pre[ry]=pre[rx];
        else pre[rx]=pre[ry];
    }
    int judge(int x,int y)
    {
        int rx=find(x),ry=find(y);
        if(rx==ry) return 1;
        return 0;
    }
    int main()
    {
        int T,M;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&N,&M);
            int x,y;
            init();
            char s[10];
            while(M--)
            {
                scanf("%s",s);
                scanf("%d%d",&x,&y);
                if(s[0]=='A')
                {
                    if(judge(x,y)||judge(x+N,y+N)) printf("In the same gang.
    ");
                    else if(judge(x,y+N)||judge(x+N,y)) printf("In different gangs.
    ");
                    else printf("Not sure yet.
    ");
                    
                }
                else
                {
                    unit(x+N,y);
                    unit(x,y+N);
                }
            }
        }
        return 0;
    }
    View Code
  • 相关阅读:
    showModalDialog和showModelessDialog使用心得
    C# 实现验证文本框中输入的是数值型??
    将Asp.Net页面输出到EXCEL里去
    timespan 和 datetime 的比较
    使用 Reporting Services 中的窗体身份验证
    DataGrid使用技巧················CSDN 收藏地址。。
    公司如何让留住技术人员??
    如何用多线程来实现ping多台机器??
    Datagrid根据选择的checkbox编辑和更新多行记录····
    做技术,切不可沉湎于技术 !
  • 原文地址:https://www.cnblogs.com/ACRykl/p/9734214.html
Copyright © 2011-2022 走看看