POJ A-Wireless Network

最新推荐文章于 2020-01-14 09:06:58 发布

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最新推荐文章于 2020-01-14 09:06:58 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zlrrrr/p/9721766.html

本文介绍了一种在地震后修复网络连接的问题解决方法，利用并查集算法来判断计算机间是否能通过直接或间接的方式进行通信。在东南亚地震后，ACM团队建立的无线网络遭受破坏，本文提出了一种算法，通过修复计算机和测试通信操作，逐步恢复网络功能。

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http://poj.org/problem?id=2236

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

题解：并查集

代码：

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

int f[1010];
int N, d;
bool rep[1010];
int q, p, n;

int st[1010];
int en[1010];

void init() {
    for(int i = 1; i <= N; i ++) {
        f[i] = i;
        rep[i] = false;
    }
}

int Find(int x) {
    if(f[x] != x) f[x] = Find(f[x]);
    return f[x];
}

void Merge(int x, int y) {
    int fx = Find(x);
    int fy = Find(y);
    if(fx != fy)
        f[fx] = fy;
}

int main() {
    scanf("%d%d", &N, &d);
    init();
    for(int i = 1; i <= N; i ++)
        scanf("%d%d", &st[i], &en[i]);

    int f1, f2;
    char op;
    while(~scanf("%c", &op)) {
        if(op == 'O') {
            scanf("%d", &n);
            rep[n] = true;
            for(int i = 1; i <= N; i ++) {
                if(i != n && rep[i] && (en[i] - en[n]) * (en[i] - en[n]) + (st[i] - st[n]) * (st[i] - st[n]) <= d * d) {
                    f1 = Find(n);
                    f2 = Find(i);
                    f[f1] = f2;
                }
            }
        }
        else if(op == 'S'){
            scanf("%d%d", &q, &p);
            f1 = Find(p);
            f2 = Find(q);

            if(f1 == f2) printf("SUCCESS\n");
            else printf("FAIL\n");
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/zlrrrr/p/9721766.html