POJ 2236 Wireless Network

Wireless Network

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 20584 Accepted: 8657

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 10 10 20 30 4O 1O 2O 4S 1 4O 3S 1 4

Sample Output

FAILSUCCESS

Source

POJ Monthly,HQM

思路：简单的并查集应用，对任意两台电脑判断能否直接或者间接连通，注意计算距离时用浮点数就行了

AC代码如下：

#include <iostream>#include <cmath>#include <cstring>#include <cstdio>using namespace std;const int maxn=1000+5;int vis[maxn];int fa[maxn];struct Point{    int x,y;}pp[maxn];int myfind(int x){    if(fa[x]!=x) return fa[x]=myfind(fa[x]);    return fa[x];}double getDis(int i,int j){    return sqrt(1.0*(pp[i].x-pp[j].x)*(pp[i].x-pp[j].x)+(pp[i].y-pp[j].y)*(pp[i].y-pp[j].y));}int main(){    int n,d;    scanf("%d%d",&n,&d);    for(int i=1;i<=n;i++){        fa[i]=i;        scanf("%d%d",&pp[i].x,&pp[i].y);    }    char x;    int a,b;    getchar();    while(scanf("%c",&x)!=EOF){        //printf("%c",x);        if(x=='O'){            scanf("%d",&a);            vis[a]=1;            for(int i=1;i<=n;i++){                int tb=myfind(a);                if(vis[i] && getDis(a,i)<=d){                    int ta=myfind(i);                    if(ta!=tb){                        fa[ta]=tb;                    }                }            }        }    else {        scanf("%d%d",&a,&b);        int ta=myfind(a);        int tb=myfind(b);        if(ta!=tb){            cout<<"FAIL"<<endl;        }        else cout<<"SUCCESS"<<endl;    }    getchar();    }    return 0;}