POJ -2236 Wireless Network

Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 12151		Accepted: 5129

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
代码如下：

/*
解题思路：
	并查集的简单应用，对每次修好的电脑对其它已经修好的电脑遍历，
	如果距离小于等于最大通信距离就将他们合并。之后判断2台电脑是不是一个集合中就KO了
*/
/*代码一：自己的代码
#include<iostream>
#include<cstring>
#include<cmath>

int point[1002][2],root[1002];
bool dis[1002][1002],work[1002];
int cal(int x1,int y1,int x2,int y2)
{
	return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}

int find_root(int i)
{
	while(root[i]>0)
		i=root[i];
	return i;
	//return root[i]<0?i:find_root(root[i]);

}

void merge(int i,int j)
{
	i=find_root(i);
	j=find_root(j);
	if(i!=j)
	{
		if(root[i]<root[j])//此时root[i]的绝对值比root[j]大
		{
			root[i]+=root[j];
			root[j]=i;
		}
		else
		{
			root[j]+=root[i];
			root[i]=j;
		}
	}
}

int main()
{
	using namespace std;
	int n,i,j,a,b,d;
	char s;
	while(cin>>n>>d)
	{
		memset(work,false,sizeof(work));
		memset(root,-1,sizeof(root));
		for(i=1;i<=n;++i)
			cin>>point[i][0]>>point[i][1];
		for(i=1;i<=n;++i)
			for(j=1;j<=i;++j)
			{
				int t=cal(point[i][0],point[i][1],point[j][0],point[j][1]);
				if(t<=d*d)
					dis[i][j]=dis[j][i]=true;
				else
					dis[i][j]=dis[j][i]=false;
			}
		while(cin>>s)
		{
			if(s=='O')
			{
				cin>>a;
				work[a]=true;
				for(i=1;i<=n;++i)
				{
					if(i!=a&&work[i]&&dis[i][a])
						merge(i,a);
				}
			}
			else if(s=='S')
			{
				cin>>a>>b;
				if(find_root(a)==find_root(b))
					cout<<"SUCCESS"<<endl;
				else
					cout<<"FAIL"<<endl;
			}
		}
	}
	return 0;
}
*/

//代码二:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define N 1110
int d;
bool use[N];

struct node
{
	int pre;
	int x, y;
}p[N];

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

void join(const node p1, const node p2)
{
	int root1, root2;
	root1 = find(p1.pre);
	root2 = find(p2.pre);
	if(root1 != root2)
		if((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y) <= d * d)
			p[root2].pre = root1;
}

int main()
{
	//freopen("Input.txt", "r", stdin);
	int num;
	char ope;
	int ok;
	int from, to;
	scanf("%d%d", &num, &d);
	for(int i = 1; i <= num; ++i)
		p[i].pre = i;
	memset(use, false, sizeof(use));
	for(int i = 1; i <= num; ++i)
		scanf("%d%d", &p[i].x, &p[i].y);
	while(scanf("\n%c", &ope) != EOF)
	{
		if(ope == 'O')
		{
			scanf("%d", &ok);
			use[ok] = true;
			for(int i = 1; i <= num; ++i)
				if(use[i] && i != ok)
					join(p[i], p[ok]);
		}
		else
		{
			scanf("%d%d", &from, &to);
			if(find(from) == find(to))
				printf("SUCCESS\n");
			else
				printf("FAIL\n");
		}
	}
	return 0;
}

 

转载于:https://www.cnblogs.com/dongsheng/archive/2012/07/21/2602065.html