zoukankan      html  css  js  c++  java
  • poj 2236 Wireless Network 【并查集】

    Wireless Network
    Time Limit: 10000MS   Memory Limit: 65536K
    Total Submissions: 16832   Accepted: 7068

    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 1
    0 1
    0 2
    0 3
    0 4
    O 1
    O 2
    O 4
    S 1 4
    O 3
    S 1 4
    

    Sample Output

    FAIL
    SUCCESS

    注意要区分修过的和没修过的

    代码:

    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #include <math.h>
    using namespace std;
    #define M 1005
    #define LL __int64
    
    struct node{
    	int x, y;
    }s[M];
    int map[M][M], fat[M], vis[M];
    
    int f(int x){
    	if(fat[x] != x) fat[x] = f(fat[x]);
    	return fat[x];
    }
    
    int main(){
    	int n, d;
    		scanf("%d%d", &n, &d);
    		int i, j;
    		d*=d;
    		for(i = 1; i <= n; i ++){
    			scanf("%d%d", &s[i].x, &s[i].y);
    			map[i][i] = 0;
    			for(j = i-1; j> 0; j --){
    				map[j][i] = map[i][j] = (s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y);
    			//	printf("%d..map(%d, %d)
    ", map[i][j], i, j);
    			}
    		}
    		char ss[2];
    		int a, b;
    		memset(vis, 0, sizeof(vis));
    		for(i = 1; i <= n; i++) fat[i] = i;
    		while(scanf("%s", ss) == 1){
    			if(ss[0] == 'O'){
    				scanf("%d", &a);
    				vis[a] = 1;
    				for(i = 1; i <= n; i ++){
    					if(vis[i]&&map[a][i] <= d&&i != a){
    						int x = f(a); int y = f(i);
    						if(x != y) fat[x] = y;
    					}
    				}
    			}
    			else{
    				scanf("%d%d", &a, &b);
    				int x = f(a); int y = f(b);
    				if(vis[a]&&vis[b]&&x == y) printf("SUCCESS
    ");
    				else printf("FAIL
    ");
    			}
    		}
    	return 0;
    }


  • 相关阅读:
    spring boot的actuator
    mongo的用户角色配置
    spring boot使用AOP统一处理web请求
    (原)luarocks install 提示 failed fetching manifest
    (原)ubuntu中安装kate
    (原+转)Ubuntu16.04软件中心闪退及wifi消失
    (原)torch使用caffe时,提示CUDNN_STATUS_EXECUTION_FAILED
    (原)torch中微调某层参数
    (原)torch的apply函数
    (原)torch的训练过程
  • 原文地址:https://www.cnblogs.com/cxchanpin/p/6780443.html
Copyright © 2011-2022 走看看