zoukankan      html  css  js  c++  java
  • 并查集+向量偏移

    http://poj.org/problem?id=1703

    Find them, Catch them
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 31589   Accepted: 9739

    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.

    题目大意:在这个城市里有两个黑帮团伙,现在给出N个人,问任意两个人他们是否在同一个团伙
    输入D x y代表xy不在一个团伙里
    输入A x y要输出xy是否在同一团伙或者不确定他们在同一个团伙里

    用图解释一下并查集的向量偏移

    加入12是敌人,23是敌人,用mark[]标记01,更新如下;

    a=1;b=2  ==>  x=1,y=2; ==> f[1]=2;

    Mark[1]=(mark[1]+mark[2]+1)%2=1;

    a=2,b=3 ==> x=2,y=3; ==> f[2]=3;

    Mark[2]=(mark[2]+mark[3]+1)%2=1;

    现在找13是什么关系;

    递归过程:

    x=1;f[x]=2;x!=f[x]; t=f[x]=2  ==<f[x]=finde(f[x])>==> 

    x=2;f[x]=3;x!=f[x], t=f[x]=3  ==<f[x]=finde(f[x])>==>

    x=3;f[x]=3; return f[x];  <==f[x]=finde(f[x])==

    Mark[2]=(mark[2]+mark[t=3])%2=1; return f[x];  <==f[x]=finde(f[x])==

    Mark[1]=(mark[1]+mark[t=2])%2=0; return f[x];

    更新完;

    发现mark[1]==mark[3];就是一类;

     程序:

    #include"stdio.h"
    #include"string.h"
    #include"queue"
    #include"stack"
    #include"math.h"
    #include"iostream"
    #define M 100005
    #define inf 100000000
    #define mod 10007
    #define eps 1e-10
    using namespace std;
    int f[M],mark[M];
    int finde(int x)
    {
        if(x!=f[x])
        {
            int t=f[x];
            f[x]=finde(f[x]);
            mark[x]=(mark[x]+mark[t])%2;
        }
        return f[x];
    }
    void make(int a,int b)
    {
        int x=finde(a);
        int y=finde(b);
        if(x!=y)
        {
            f[x]=y;
            mark[x]=(mark[a]+mark[b]+1)%2;
        }
    }
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int n,m,i;
            scanf("%d%d",&n,&m);
            for(i=1;i<=n;i++)
                f[i]=i;
            memset(mark,0,sizeof(mark));
            while(m--)
            {
                int a,b;
                char str[3];
                scanf("%s%d%d",str,&a,&b);
                if(str[0]=='D')
                {
                    make(a,b);
                }
                else
                {
                    int x=finde(a);
                    int y=finde(b);
                    if(x!=y)
                        printf("Not sure yet.
    ");
                    else if(mark[a]==mark[b])
                        printf("In the same gang.
    ");
                    else
                        printf("In different gangs.
    ");
                }
            }
        }
        return 0;
    }
    



  • 相关阅读:
    HTML style基础2
    HTML 笔记 基础1
    qq发送邮件
    Requests+Excel接口自动化测试(Python)
    echars前端处理数据、pyechars后端处理数据
    appium基础一:连接手机和appium-desktop定位元素
    Loadrunner 性能测试工具笔记
    总结windows cmd 查看进程,端口,硬盘信息
    appium + python 自动化调试手机时 UiAutomator exited unexpectedly with code 0, signal null
    什么是接口测试?怎样做接口测试?
  • 原文地址:https://www.cnblogs.com/mypsq/p/4348175.html
Copyright © 2011-2022 走看看