poj Wireless Network(基础并查集)
来源:互联网 发布:淘宝上传宝贝失败 编辑:程序博客网 时间:2024/06/05 21:50
Wireless Network
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.
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.
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
代码:
#include<stdio.h>#include<math.h>#include<string.h>#define maxn 1000+10int pre[maxn],xi[maxn],yi[maxn],vis[maxn];int n;double d;void init(){ for(int i=1; i<=n; i++) pre[i]=i;}int Find(int x){ int r=x; while(pre[r]!=r) r=pre[r]; int i=x,j; while(i!=r) { j=pre[i]; pre[i]=r; i=j; } return r;}void join(int x,int y){ int fx=Find(x),fy=Find(y); if(fx!=fy) pre[fx]=fy;}int check(int x1,int y1,int x2,int y2){ if(d*d>=((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))) return 1; return 0;}int main(){ int i,j; scanf("%d%lf",&n,&d); init(); for(i=1; i<=n; i++) scanf("%d%d",&xi[i],&yi[i]); char ss[10]; while(~scanf("%s",ss)) { int x,y; if(ss[0]=='O') { scanf("%d",&x); vis[x]=1; for(i=1; i<=n; i++) if(vis[i]&&check(xi[i],yi[i],xi[x],yi[x])&&x!=i) join(i,x); } else { scanf("%d%d",&x,&y);// for(i=1;i<=n;i++)// printf("pre[%d]=%d\n",i,pre[i]); if(Find(x)==Find(y))//注意这里,目前的pre[x]并不一定等于目前的pre[y],所以用Find[] printf("SUCCESS\n"); else printf("FAIL\n"); } } return 0;}
ps:错误地认为pre[x]==pre[y]就是判断结果的条件,谨记
1 0
- poj Wireless Network(基础并查集)
- poj Wireless Network(基础并查集)
- POJ 2236 Wireless Network(并查集基础)
- POJ Wireless Network (并查集)
- POJ 2236 Wireless Network(并查集)
- POJ 2236 Wireless Network //并查集
- POJ 2236 WIRELESS NETWORK(并查集)
- POJ-2236 wireless network 并查集
- poj 2236 Wireless Network(并查集)
- poj 2236 - Wireless Network(并查集)
- POJ P2236 Wireless Network 并查集
- POJ 2236 Wireless Network(并查集)
- poj 2236 Wireless Network(并查集)
- poj 2236 Wireless Network(并查集)
- poj 2236 Wireless Network 并查集
- POJ 2236 Wireless Network 并查集
- poj 2236 Wireless Network 并查集
- poj 2236 Wireless Network 【并查集】
- 魅力物理2
- 安装Android Studio 可能遇到的问题及解决办法
- 基础查询
- 杭电OJ 2009.求数列的和
- 74. Search a 2D Matrix
- poj Wireless Network(基础并查集)
- 使用JDK开发WebService-客户端建立
- 使用TOP查询
- uva 10334Ray Through Glasses
- RTMP学习(八)rtmpdump源码阅读(3)初始化与URL解析
- leecode 解题总结:324. Wiggle Sort II
- (1)第一次产品面试的感受
- Android面试题集
- 蓝桥杯2012年试题微生物增值