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Wireless Network

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

Output


FAIL

SUCCESS

苦逼大学生又要打开发又要打算法还要准备考试考证（发个牢骚 生活还是要继续）

并查集模版题，而且找相邻点的时候暴的过去，计算距离的时候不要开根同时平方就好了，精度问题会导致WA

#include <iostream>

using namespace std;

const int N = 1e3 + 5;
typedef pair<int, int> PII;

int p[N];
PII point[N];
int n, d;
bool vis[N]; // 判断这个点修好了没

void init(int x) {
    for(int i = 1; i <= x; ++i) {
        p[i] = i;
    }
}

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

void merge(int a, int b) {
    p[find(a)] = find(b);
}
//以上为并查集模板 建议全文背诵

bool dis(int a, int b) {
    if((point[a].first - point[b].first) * (point[a].first - point[b].first) + (point[a].second - point[b].second) * (point[a].second - point[b].second) <= d * d)
        return 1;
    return 0;
}

int main() {

    cin >> n >> d;
    init(n);

    for(int i = 1; i <= n; ++i) {
        cin >> point[i].first >> point[i].second;
    }

    string s;

    while(cin >> s) {

        if(s == "O") {
            int a;
            cin >> a;
            vis[a] = 1;

            int fa = find(a);

            for(int i = 1; i <= n; ++i) {
                if(vis[i] && fa != find(i) && dis(i, a)) {
                    merge(a, i);
                }
            }
        } else {
            int a, b;
            cin >> a >> b;

            if(find(a) == find(b))
                cout << "SUCCESS" << endl;
            else 
                cout << "FAIL" << endl;
        }
    }



    return 0;
}