【POJ 2236 Wireless Network】 + 并查集

Wireless Network
Time Limit: 10000MS Memory Limit: 65536K
Total Submissions: 25271 Accepted: 10496

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

Sample Output

FAIL
SUCCESS

题意 ： 给出电脑个数 N,可以联网的最大距离 D，N 个电脑的位置坐标，起初电脑都是怀的，有未知次询问，如果为 ‘O’ 询问，第 i 台电脑被修好，’S’询问第 x 和 y台电脑能否可以联网

思路 ： 并查集，一个集合里的代表可以联网，每修好一台电脑后，即遍历其他修好的电脑，若距离在联网范围内者和为一个集合～

AC代码：

#include<cstdio>#include<cstring>using namespace std;char st[2];int vis[1010],f[1010],dx[1010],dy[1010],D;int bc(int x){    int p = x;    while(x != f[x]) // 寻找父节点        x = f[x];    while(p != x){ // 路径压缩，出于习惯一般会加上        int t = f[p];        f[p] = x;        p = t;    }    return x;}int dis(int x,int y){    int vx = dx[x] - dx[y],vy = dy[x] - dy[y];    if(vx * vx + vy * vy <= D * D)        return 1;    else        return 0;}int main(){    int N,x,y;    scanf("%d %d",&N,&D);    for(int i = 1 ; i <= N; i++)        f[i] = i;    for(int i = 1 ; i <= N; i++)        scanf("%d %d",&dx[i],&dy[i]);    while(scanf("%s",st) != EOF){        if(st[0] == 'O'){            scanf("%d",&x);            vis[x] = 1;            for(int i = 1 ; i <= N; i++){ // 把可以‘联网’的电脑化为一个集合                if(i != x && vis[i] && dis(x,i)){                    int fx = bc(x),fi = bc(i);                    f[fx] = fi;                }            }        }        else{            scanf("%d %d",&x,&y);            int fx = bc(x),fy = bc(y);            if(fx == fy)                printf("SUCCESS\n");            else                printf("FAIL\n");        }    }    return 0;}