Wireless Network（并查集）

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本文介绍了一个模拟地震后无线网络修复的算法问题。通过并查集数据结构实现对计算机网络的逐步修复，解决设备间的通讯判断问题。

Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 14086		Accepted: 5962

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

Sample Output

FAIL
SUCCESS

题意：

有N（1到1001）台机器和D最小距离（0到20000），给出每台机器的坐标（X,Y）（0到10000）。O代表已经修好某台机器，S代表询问A和B机器间是否能联系上。要能联系则必须两机器的坐标距离不大于D，如果A和B能连上，B和C能连上，那么A和C也能连上。输出当询问S时，两部机器能联系上的结果，能则输出SUCCESS，不能则输出FAIL。

思路：

普通并查集。但是与一般不同是它有一个距离的条件，并且每次输入O可能会更新一次连通集合。用rep数组来表示机器i是否已经修好，1为修好，0为没有修好。当输入O时，循环一次所有机器，当循环到有一台机器满足距离小于1，且这部机器已经是修好的，那就将该机器合并到该集合中。当输入S时，判断A和B的根节点是否相同则知道是否能连接上。

AC：

#include<stdio.h>
#include<math.h>
#include<string.h>
typedef struct 
{
	int x;
	int y;
}pos;

pos num[1005];
int root[1005],rep[1005];
int n;

double distance(pos a,pos b)
{
    double s;
    s=sqrt(pow((double)a.x-(double)b.x,2)+pow((double)a.y-(double)b.y,2));
    return s;
}

int find(int a)
{
	int x=a,t;
	while(x!=root[x])
	   x=root[x];
	while(x!=a)
	{
		t=root[a];
		root[a]=x;
		a=t;
	}
	return x;
}

void merge(int a,int b)
{
	int fa,fb;
	fa=find(a);
	fb=find(b);
	if(fa!=fb) root[fa]=fb;
}

int main()
{
	int d;
	char c[5];
	int a,b;
	scanf("%d%d",&n,&d);
	memset(rep,0,sizeof(rep));
	for(int i=1;i<=n;i++)
	    root[i]=i;
	
	for(int i=1;i<=n;i++)
	 	scanf("%d%d",&num[i].x,&num[i].y);
	
	while(scanf("%s",c)!=EOF)
	{
		if(c[0]=='O')
		{
			scanf("%d",&a);
			rep[a]=1;
			for(int i=1;i<=n;i++)
			  if(distance(num[i],num[a])<=d&&rep[i])  merge(a,i);
//这里的距离因为看了样例，然后固化了，写成小于1.0，然后又WA了N次
		}
		else
		{
			scanf("%d%d",&a,&b);
			if(find(a)==find(b)) printf("SUCCESS\n");
			else printf("FAIL\n");
//这里一开始写成FALL，WA了一次
		}
	}
	return 0;
}

总结：

1.一开始定义输入字符char c，输入的时候scanf(“%c”，&c)出现了错误，输入scanf(“% c”，&c)就不会出现错误了，也就是输入格式前面加个空格；

2.FAIL一开始打成FALL，低级错误；

3.距离是小于d不是小于样例给的1，又是低级错误；

4.一开始想的是：当O的时候登记rep为修好状态，当S的时候再一个个匹配合并，然后用了一个二重循环，先找到一个已经修好的，然后以这个修好的为基准再循环一层来找下一个修好的，然后匹配。这样无疑会TLE的。改进后就是每输入一次O就更新一次连通集合，没输入一次S就判断一次两者是否属于一个集合就可以了。

5.一开始还想过用一个二维数组来保存i到j是否能连上的状态，其实不需要，改进后只需要用个rep数组登记就好。