POJ - 2236Wireless Network-並查集

標籤：url   word   通訊   nes   tracking   idg   back   pad   set   

id=11125" target="_blank" style="color:blue; text-decoration:none">POJ - 2236 id=11125" target="_blank" style="color:blue; text-decoration:none">Wireless Network Time Limit: 10000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u id=11125" class="login ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="display:inline-block; position:relative; padding:0px; margin-right:0.1em; vertical-align:middle; overflow:visible; text-decoration:none; font-family:Verdana,Arial,sans-serif; border:1px solid rgb(211,211,211); color:blue; font-size:12px!important">Submit Status 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 首先要看懂題目的意思。首先要進行通訊，首先你的電腦必須已經修好，否則是不可以通訊的，並且通訊時距離在一定範圍內才可以實現 /*Author: 2486Memory: 380 KBTime: 3063 MSLanguage: G++Result: Accepted*/#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int maxn=10000+5;int N,d;char op[10];int par[maxn],ranks[maxn];bool vis[maxn];//代表著是否已經修複了電腦，電腦是否正常struct co { int x,y;} coh[maxn];void init(int sizes) { for(int i=0; i<=sizes; i++) { par[i]=i; ranks[i]=1; }}int find(int x) { return par[x]==x?x:par[x]=find(par[x]);}bool same(int x,int y) { return find(x)==find(y);}bool judge(int x,int y) { return pow((coh[x].x-coh[y].x),2.0)+pow((coh[x].y-coh[y].y),2.0)<=pow(d,2.0);//推斷距離是否符合條件}void unite(int x,int y) { x=find(x); y=find(y); if(x==y)return ; if(ranks[x]>ranks[y]) { par[y]=x; } else { par[x]=y; if(ranks[x]==ranks[y])ranks[x]++; }}int main() { int c,e; scanf("%d%d",&N,&d); init(N); memset(vis,false,sizeof(vis)); for(int i=1; i<=N; i++) { scanf("%d%d",&coh[i].x,&coh[i].y); } while(~scanf("%s",op)) { if(op[0]=='O') { scanf("%d",&c); vis[c]=true; for(int i=1; i<=N; i++) { if(vis[i]&&judge(c,i)) {//與其它的電腦進行聯絡的前提是他們是好的，假設不好就串連不上 unite(c,i); } } } else { scanf("%d%d",&c,&e); if(vis[c]&&vis[e]&&same(c,e)) {//是否通訊。要看電腦是否已經修複完畢 printf("SUCCESS\n"); } else printf("FAIL\n"); } } return 0;}

POJ - 2236Wireless Network-並查集