POJ-2236 Wireless Network（并查集）

最新推荐文章于 2021-10-25 19:11:05 发布

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#ACM #并查集 #Wireless Network #POJ #无线网络

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本文介绍了一种在地震后的网络修复算法，通过计算计算机间距离来判断是否可以建立连接，并使用并查集来维护网络状态。

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Problem 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:

Sample Output:

FAIL
SUCCESS

思路：

修理一台电脑就是进行想要进行合并操作，能不能合并还要看与其他已经被修好的电脑的距离，距离小于等于D才可以进行合并。查询时直接调用Find函数。

上AC代码：

#include <stdio.h>
#include <math.h>
#include <string.h>
int f[1002];
typedef struct
{
    int x;
    int y;
}point;
point pos[1002];
bool beRepair[1002];
int n,d;
char str[6];
//Find函数
int Find(int x)
{
    if(x==f[x])
        return x;
    else
        return f[x]=Find(f[x]);
}
int main()
{
    int i;
    scanf("%d%d",&n,&d);
    for(i=1;i<=n;i++)
    {
        f[i]=i;
        scanf("%d%d",&pos[i].x,&pos[i].y);
    }
    memset(beRepair,0,sizeof(beRepair));
    while(~scanf("%s",str))
    {
        if(str[0]=='O')
        {
            int id;
            //修理操作
            scanf("%d",&id);
            beRepair[id]=true;
            for(i=1;i<=n;i++)
            {
                if(id==i)
                    continue;
                if(beRepair[i]==true)
                {
                    if(sqrt((pos[id].x-pos[i].x)*(pos[id].x-pos[i].x)+(pos[id].y-pos[i].y)*(pos[id].y-pos[i].y))<=d)
                    {
                        f[Find(i)]=Find(id);
                    }
                }
            }
        }
        if(str[0]=='S')
        {
            //查询操作
            int a,b;
            scanf("%d%d",&a,&b);
            if(Find(a)==Find(b))
            {
                printf("SUCCESS\n");
            }
            else
            {
                printf("FAIL\n");
            }
        }
    }
    return 0;
}