POJ 2236 Wireless Network 【并查集的简单应用 判断是否在同一连通分量】

Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 13361		Accepted: 5659

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

Source

POJ Monthly,HQM

题意：给你 N 台坏了的电脑的坐标 ，和一个距离范围 d .

(在距离范围内的电脑可以相互通信，每台电脑也可以连接两台不同的电脑，使他们之间能够通信)

输入任意次数操作：

O x 表示修理好编号为 x 台的电脑

S x y 判断电脑 x 和 y 是否能够通信 。能则 SUCCESS ，不能则 FAIL

算法：并查集简单应用，判断两点是否在同一连通分量

思路：并查集部分很好处理。

关键是判断两台电脑是否在给定范围内，这里要用一个map[][]数组标记连通性，用一个used[]数组标记是否修理过。

每修一台电脑，则连接【合并】这台电脑和已经距离这台电脑 d 范围内的已经修好了的电脑所在的集合即可。

H

Accepted

1164 KB

1125 ms

C++

1714 B

2013-04-10 18:48:34

#include<cstdio>
#include<cmath>
#include<cstring>

const double esp = 1e-7;
const int maxn = 1000+10;

int p[maxn]; //父节点
bool used[maxn]; //标记改点是否修理
bool map[maxn][maxn]; //标记连通性

struct Point
{
double x, y;
}point[maxn];

int find(int x)
{
return x == p[x] ? x : p[x] = find(p[x]);
}

void Union(int x, int y)
{
int fx = find(x);
int fy = find(y);

if(fx == fy) return;

p[fx] = fy; //随意
}
double dist(Point a, Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main()
{
int n, d;
while(scanf("%d%d", &n, &d) != EOF)
{
memset(map, false, sizeof(map));
memset(used, false, sizeof(used));

for(int i = 1; i <= n; i++)
{
scanf("%lf%lf", &point[i].x, &point[i].y);
}

for(int i = 1; i <= n; i++)
{
p[i] = i;
map[i][i] = true;
for(int j = i+1; j <= n; j++)
{
if(dist(point[i], point[j])-d < esp)
{
map[i][j] = true; //在距离范围内
map[j][i] = true;
}
}
}

char c;
int x, y;
getchar();
while(scanf("%c", &c) != EOF)
{
if(c == 'O')
{
scanf("%d%*c", &x);
if(!used[x]) //以前未被修理
{
for(int i = 1; i <= n; i++) //依次判断
{
if(used[i] && map[i][x]) Union(x, i);
}
used[x] = true;
}
}
else if(c == 'S')
{
scanf("%d%d%*c", &x, &y);
if(find(x) == find(y)) printf("SUCCESS\n"); //连通
else printf("FAIL\n"); //不连通
}
}
}
return 0;
}