POJ：2236-Wireless Network

Wireless Network

Time Limit: 10000MS Memory Limit: 65536K
Total Submissions: 34265 Accepted: 14222

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

---

解题心得：

1. 题意就是有n台电脑，每台电脑有一个坐标，它只可以联系和它距离在d内的电脑，现在每台电脑都是坏的，有两种操作，第一种就是维修某台电脑，第二种就是询问两台电脑是否可以通信。
2. 时间给得很足啊，整整十秒，也就很简单了，每次暴力维护可以保持相互通信的电脑

---

#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1010;

int father[maxn],n,d;
bool vis[maxn];

struct Pos {
    int x,y;
}p[maxn];

void init() {
    scanf("%d%d",&n,&d);
    for(int i=1;i<=n;i++)
        father[i] = i;
    for(int i=1;i<=n;i++)
        scanf("%d%d",&p[i].x,&p[i].y);
}

int find(int x) {
    if(father[x] == x)
        return x;
    return father[x] = find(father[x]);
}

void merge(int x,int y) {
    int fx = find(x);
    int fy = find(y);
    father[fx] = fy;
}

void fix(int pos) {
    vis[pos] = true;
    for(int i=1;i<=n;i++) {
        if(vis[i]) {
            int d2 = (p[i].x-p[pos].x)*(p[i].x-p[pos].x) + (p[i].y-p[pos].y)*(p[i].y-p[pos].y);
            if(d2 <= d*d) {
                if(find(i) != find(pos)) {
                    int fi = find(i);
                    for(int j=1;j<=n;j++) {
                        if(father[j] == fi)
                            merge(j,pos);
                    }
                }
            }
        }
    }
}

int main(){
    init();
    char s[5];
    while(scanf("%s",s) != EOF) {
        int pos,a,b;
        if(s[0] == 'O') {
            scanf("%d",&pos);
            if(vis[pos])
                continue;
            else
                fix(pos);
        } else {
            scanf("%d%d",&a,&b);
            if(vis[a] && vis[b] && find(a) == find(b))
                printf("SUCCESS
");
            else
                printf("FAIL
");
        }
    }
    return 0;
}