Poj-2236 Wireless Network

题目直通车 POJ-2236

知识要点：并查集

Wireless Network

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 22625 Accepted: 9491

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 3
0 2
0 4
O 4
O 1O 2
S 1 4
S 1 4
O 3

Sample Output

FAIL
SUCCESS

题目意思：

大致上就是地震之后有N 台电脑损坏，现在想要创建无线网络，无线网络能够通讯的长度为d.当输入为 O p 时，修理电脑p;当输入为 S p q 时候，询问p，q能否联通。

样例分析：

4 台电脑，无线通讯距离为1，  四台电脑坐标分别为 (0,1)  (0,2)  (0,3) (0,4) . 然后修理电脑 1,2,4 (那么此时1,2能够互相通讯,4 独立)  当修理电脑3时,(1,2,3,4)都能相互通讯。

代码：

#include <stdio.h>#include <math.h>int pre[3000];int find (int x){int r = x;while (pre[r]!=r){r = pre[r];                 // 找到 x 的最上级}int i = x,j;while (pre[i]!=r)                   // 压缩路径， 将并查集压缩为只有两级的树（根节点，叶节点）{j = pre[i];pre[i] = r;i = j;}return r;}void mix(int x,int y)                       // 将 x，y接通，成为一个集合{int fx = find(x),fy = find(y);if(fx!=fy){pre[fy]=fx;}}struct computer{int x;int y;}com[3000];int main(){int N,a,i,x,y;char c;int rep[3000];int repcont=1;scanf("%d%d",&N,&a);for(i=1;i<=N;i++){scanf("%d%d",&com[i].x,&com[i].y);}for(i=0;i<=N;i++){pre[i]=i;                      // 初始化前驱数组}while (scanf("%c",&c)!=EOF){if(c=='O'){scanf("%d",&x);rep[repcont++]=x;       // 将修好的电脑存入数组for(i=1;i<repcont;i++)  // 遍历已修好的电脑{if((sqrt(pow((com[rep[i]].x-com[x].x),2)+pow((com[rep[i]].y-com[x].y),2)))<=a)mix(rep[i],x);                                //如果当前电脑和已修好的电脑能够互通，那么混合}}else if(c=='S') {scanf("%d%d",&x,&y);if(find(x)==find(y))                                 {printf("SUCCESS\n");}else printf("FAIL\n");}}return 0;}