poj 2236:Wireless Network（並查集，提高題）

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Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 16065		Accepted: 6778

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 　　 並查集，提高題。　　算是並查集的應用，拐了點彎。 　　 題意：　　有一個電腦網路的所有線路都壞了，網路中有n台電腦，現在你可以做兩種操作，修理（O）和檢測兩台電腦是否連通（S），只有修理好的電腦才能連通。連通有個規則，兩台電腦的距離不能超過給定的最大距離D（一開始會給你n台電腦的座標）。檢測的時候輸出兩台電腦是否能連通。 　　 思路：　　每次修理好一台電腦的時候就遍曆一下所有修好的電腦，看距離是否<=D，如果符合說明可以連通，將兩台電腦所在集合合并。　　每次檢查的時候判斷一下這兩台電腦是否在同一集合中即可。 　　 注意：
　　1、座標之後給出的電腦編號都是n+1的。例如O 3，他實際上修理的是編號為2的電腦，因為電腦是從0開始編號的。　　2、比較距離的時候注意要用浮點數比較，否則會WA。　　3、"FAIL"不要寫成"FALL"。　　4、字串輸入的時候注意處理好斷行符號，空格等情況。　　5、注意N的範圍（1 <= N <= 1001），最大是1001，不是1000。是個小坑，數組開小了可能會錯哦。 　　另外測試了一下使用路徑壓縮，和不使用路徑壓縮的效率。使用是1079MS（C++），不使用是2032MS（C++），時間縮短了一倍。　　另外C++和G++提交時間相差也很大。同樣的代碼G++就變成了3063MS（G++）。 　　 代碼：

 1 #include <iostream> 2 #include <stdio.h> 3 #include <cmath> 4 using namespace std; 5  6 #define MAXN 1010 7  8 int dx[MAXN],dy[MAXN];    //座標 9 int par[MAXN];    //par[x]表示x的父節點10 int repair[MAXN] ={0};11 int n;12 13 void Init()    //初始化14 {15     int i;16     for(i=0;i<=n;i++)17         par[i] = i;18 }19 20 int Find(int x)    //查詢x的根節點並路徑壓縮21 {22     if(par[x]!=x)23         par[x] = Find(par[x]);24     return par[x];25 }26 27 void Union(int x,int y)    //合并x和y所在集合28 {29     par[Find(x)] = Find(y);30 }31 32 int Abs(int n)33 {34     return n>0?n:-n;35 }36 37 double Dis(int a,int b)38 {39     return sqrt( double(dx[a]-dx[b])*(dx[a]-dx[b]) + (dy[a]-dy[b])*(dy[a]-dy[b]) );40 }41 42 int main()43 {44     int d,i;45 46     //初始化47     scanf("%d%d",&n,&d);48     Init();49 50     //輸入座標51     for(i=0;i<n;i++){52         scanf("%d%d",&dx[i],&dy[i]);53     }54     55     //操作56     char cmd[2];57     int p,q,len=0;58     while(scanf("%s",cmd)!=EOF){59         switch(cmd[0]){60         case ‘O‘:61             scanf("%d",&p);62             p--;63             repair[len++] = p;64             for(i=0;i<len-1;i++)    //遍曆所有修過的電腦，看能否聯通65                 if( repair[i]!=p && Dis(repair[i],p)<=double(d) )66                     Union(repair[i],p);67             break;68         case ‘S‘:69             scanf("%d%d",&p,&q);70             p--,q--;71             if(Find(p)==Find(q))    //是否有路72                 printf("SUCCESS\n");73             else 74                 printf("FAIL\n");75         default:76             break;77         }78     }79 80     return 0;81 }

 

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