Wireless Network-CSDN博客

本文链接：https://blog.csdn.net/CS33sun/article/details/79122285

本文介绍了一种模拟地震后无线网络修复过程的方法。通过并查集算法和距离限制，实现计算机间通讯状态的实时更新与查询，确保网络逐步恢复正常。

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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

并查集加一个距离限制，每次执行O操作的时候就把新的合并，合并中需要判断一下距离，S操作就是普通的并查集了。

#include <stdio.h>
#include <string.h>
struct node{
	int p;
	int x, y;
}f[1005];
int book[1005];
int N, d;
void init() {
	for(int i = 1; i <= N; i++) {
		f[i].p = i;
	}
}
int getf(int v) {
	if(f[v].p != v) {
		f[v].p = getf(f[v].p);
	}
	return f[v].p;
}
int getnum(node s1, node s2) {
	return (s1.x-s2.x)*(s1.x-s2.x) + (s1.y-s2.y)*(s1.y-s2.y);
}
void merge(node u, node v) {
	int t1 = getf(u.p);
	int t2 = getf(v.p);
	if(t1 != t2) {
		if(getnum(u, v) <= d*d) {
			f[t1].p = t2;
		}
	}
}
int main() {
	int q, p;
	scanf("%d %d", &N, &d);
	init();
	memset(book, 0, sizeof(book));
	for(int i = 1; i <= N; i++) {
		scanf("%d %d", &f[i].x, &f[i].y);
	}
	char op;
	while(~scanf("\n%c ", &op)) {
		if(op == 'O') {
			scanf("%d", &p);
			for(int i = 1; i <= N; i++) {
				if(book[i]) {
					merge(f[i], f[p]);
				}
			}
			book[p] = 1;
		}
		else if(op == 'S') {
			scanf("%d %d", &p, &q);
			merge(f[p], f[q]);
			int t1 = getf(p);
			int t2 = getf(q);
			if(t1 == t2) printf("SUCCESS\n");
			else printf("FAIL\n");
		}
	}
	return 0;
}