Description
An earthquake takes place in southeast Asia. The ACM (Asia cooperated Medical team) has set up a wireless network with the lap computers, but an unexpected aftershock Attacked, all computers on the network were all broken. The computers is 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 is not farther tha n d meters from it. But every computer can be regarded as the intermediary of the communication between and the other computers, that's to say Co Mputer A and Computer B can communicate if computer A and computer B can communicate directly or there is A computer C tha T can communicate with both A and B.
In the process of repairing the network, workers can take the other kinds of operations at every moment, repairing a computer, O R Testing if computers can communicate. Your job is to answer all the testing operations.
The main idea: An earthquake destroyed all the computers (should I say the power of earthquake?) Or the computer is too ruo ... Give the coordinates of each computer, two computers can be connected, when the two computers can be directly connected or have an intermediary computer can be connected to the intermediate computer directly. Two kinds of operation: O ——— repair the computer, s--query two computers can connect, output "SUCCESS", otherwise output "FAIL".
Idea: preprocessing the distance between computers, simple and check set, online practice, each repair a computer should be poor lift all the distance is less than equal to the computer, and update.
(Distance is Euler distance, with the square to save the good, notice D to First square)
Code
#include <iostream> #include <cstdio>using namespace Std;int dis[1010][1010]={0},ad[1010][2]={0},fa[1010 ]={0};bool visit[1010]={false};int rool (int x) {if (fa[x]!=x) Fa[x]=rool (fa[x]); return fa[x];} int main () {int N,i,j,r1,r2,d,a,b;char kind;cin>>n>>d;d=d*d;for (i=1;i<=n;++i) fa[i]=i;for (i=1; I<=n;++i) { scanf ("%d%d", &ad[i][0],&ad[i][1]); for (j=1;j<i;++j) dis[j][i]= Dis[i][j]= (Ad[i][0]-ad[j][0]) * (ad[i][0]-ad[j][0]) + (Ad[i][1]-ad[j][1]) * (ad[i][1]-ad[j][1]); }while (scanf ("%*c%c", &kind) ==1) {if (kind== ' O ') { scanf ("%d", &a); r1=rool (a); for (i=1;i<=n;++i) if (Dis[i][a]<=d&&visit[i]) { r2=rool (i); & nbsp if (R1!=R2) fa[r2]=r1; }  visit[a]=true;} ELSE{SCANF ("%d%d", &a,&b); R1=rool (a); R2=rool (b); if (R1!=R2) printf ("%s\n", "FAIL"), Else printf ("%s\n" , "SUCCESS");}}}
POJ2236 Wireless Network (easy)
