POJ 2236 Wireless Network(带权并查集）

旧城等待， 2022-06-17 02:23 114阅读 0赞

> Wireless Network  
> Time Limit: 10000MS Memory Limit: 65536K  
> Total Submissions: 27063 Accepted: 11205  
> Description
> 
> An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers are repaired one by one, and the network gradually began to work again. Because of the hardware restricts, each computer can only directly communicate with the computers that are not farther than d meters from it. But every computer can be regarded as the intermediary of the communication between two other computers, that is to say computer A and computer B can communicate if computer A and computer B can communicate directly or there is a computer C that can communicate with both A and B.
> 
> In the process of repairing the network, workers can take two kinds of operations at every moment, repairing a computer, or testing if two computers can communicate. Your job is to answer all the testing operations.  
> Input
> 
> The first line contains two integers N and d (1 <= N <= 1001, 0 <= d <= 20000). Here N is the number of computers, which are numbered from 1 to N, and D is the maximum distance two computers can communicate directly. In the next N lines, each contains two integers xi, yi (0 <= xi, yi <= 10000), which is the coordinate of N computers. From the (N+1)-th line to the end of input, there are operations, which are carried out one by one. Each line contains an operation in one of following two formats:  
> 1. “O p” (1 <= p <= N), which means repairing computer p.  
> 2. “S p q” (1 <= p, q <= N), which means testing whether computer p and q can communicate.  
> The input will not exceed 300000 lines.  
> Output
> 
> For each Testing operation, print “SUCCESS” if the two computers can communicate, or “FAIL” if not.  
> Sample Input  
> 4 1  
> 0 1  
> 0 2  
> 0 3  
> 0 4  
> O 1  
> O 2  
> O 4  
> S 1 4  
> O 3  
> S 1 4  
> Sample Output
> 
> FAIL  
> SUCCESS

> 题意：由n台电脑组成，因地震全部瘫痪，现在进行修复，规定距离小于等于d的电脑修复之后是可以直接相连，进行若干操作，O a，修复编号为a的电脑，S a,b 询问a,b电脑能否联系

> 题解：并查集

> 代码：

#include <iostream>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <string>
    #include <cmath>
    using namespace std;
    const int maxn = 20000+5;
    
    struct node
    {
        double xi;
        double yi;
    }a[maxn];
    int pre[maxn];//数组pre[]记录了每个点的前导点
    bool vis[maxn];//用于标记独立块的根结点
    double dis(int i,int j)
    {
        return sqrt((a[i].xi-a[j].xi)*(a[i].xi-a[j].xi)+(a[i].yi-a[j].yi)*(a[i].yi-a[j].yi));
    }
    int find(int x)
    {
        int r = x;
        while(r != pre[r])
            r = pre[r];
        int i = x,j;
        while(pre[i] != r)//路径压缩
        {
            j = pre[i];// 在改变上级之前用临时变量 j 记录下他的值
            pre[i] = r; //把上级改为根节点
            i = j;
        }
        return r;
    }
    void mix(int x,int y)
    {
        int fx = find(x);
        int fy = find(y);
        if(fx != fy)
            pre[fy] = fx;
    }
    int main()
    {
        int n,d;
        cin>>n>>d;
        memset(vis,false,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            cin>>a[i].xi>>a[i].yi;
            pre[i] = i;
        }
        string s;
        int _x,_y;
        while(cin>>s)
        {
            if(s[0] == 'O')
            {
                cin>>_x;
                vis[_x] = true;
                for(int i=1;i<=n;i++)
                {
                    if(dis(_x,i) <= d&&vis[i])
                        mix(i,_x);
                }
            }
            if(s[0] == 'S')
            {
                cin>>_x>>_y;
                if(find(_x) == find(_y))
                    cout<<"SUCCESS"<<endl;
                else cout<<"FAIL"<<endl;
            }
        }
        return 0;
    }

> 并查集模板：

int pre[maxn];//数组pre[]记录了每个点的前导点 
    int find(int x)
    {
        int r = x;
        while(r != pre[r])
            r = pre[r];
        int i = x,j;
        while(pre[i] != r)//路径压缩
        {
            j = pre[i];// 在改变上级之前用临时变量 j 记录下他的值
            pre[i] = r; //把上级改为根节点
            i = j;
        }
        return r;
    }
    void mix(int x,int y)
    {
        int fx = find(x);
        int fy = find(y);
        if(fx != fy)
            pre[fy] = fx;
    }
    int main()
    {
       ...
    }