POJ 2236 Wireless Network

最新推荐文章于 2020-09-19 15:05:29 发布

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本文探讨了在东南亚地震后，ACM医疗团队如何重建无线网络，解决计算机间通信问题。通过并查集算法，确保即使硬件受限，也能使距离不超过特定范围的计算机进行直接或间接通信。文章详细描述了输入输出格式，以及样例输入输出，帮助读者理解如何在修复过程中判断两台计算机是否能通信。

题目描述

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：并查集应用，注意如果修好了1，2，4，再修3的时候，1，2，3，4连成一片，不要只把第三个并到1，2

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

int n,d;
int tree[1020];

struct coordinate{
    int x,y;
};
coordinate coo[1020];

int findRoot(int x){
    if(tree[x]==x)return x;
    else return tree[x]=findRoot(tree[x]);
}

int close(int &a,int &i){
    if( (coo[a].x-coo[i].x)*(coo[a].x-coo[i].x) + (coo[a].y-coo[i].y)*(coo[a].y-coo[i].y) > d*d)
        return 0;
    else return 1;
}

void deal(int &a){
    int flag=0;
    for(int i=1;i<=n;i++){
        if(i!=a&&tree[i]&&close(a,i)){
            int ri=findRoot(i);
            if(!flag){
                tree[a]=ri;   //把a并到别处
                flag=1;
            }
            else {
                tree[ri]=findRoot(a);
            }
        }
    }
}

int main()
{
    scanf("%d%d",&n,&d);
    memset(tree,0,sizeof(tree));
    for(int i=1;i<=n;i++){
        scanf("%d%d",&coo[i].x,&coo[i].y);
    }

    char ch;
    while(~scanf("%c",&ch)){
        if(ch=='O'){
            int a;
            scanf("%d",&a);
            tree[a]=a;
            deal(a);
        }
        if(ch=='S'){
            int a,b;
            scanf("%d%d",&a,&b);
            int ra=findRoot(a);
            int rb=findRoot(b);
            if(ra&&rb&&ra==rb){
                printf("SUCCESS\n");
            }
            else {
                printf("FAIL\n");
            }
        }
    }
    return 0;
}