ACM并查集Wireless Network

最新推荐文章于 2022-06-13 12:19:46 发布

原创最新推荐文章于 2022-06-13 12:19:46 发布 · 208 阅读

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本文介绍了一种使用并查集算法解决计算机网络修复问题的方法。在地震导致网络中断后，通过逐步修复计算机节点，并检查节点间是否能建立通信。文章详细阐述了如何利用并查集进行节点连接状态的维护及查询。

题目：

n 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

题目大意：给出各个点的坐标，以及能够取得联系的距离，判断某两个点之间能否联系

很明显的一个并查集问题，只是输入让我愣了一会，重点认为是输入吧，跟模板相比其实只是多了一个能否联系的判断

代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int f[1005],book[1005];
int n,d;
struct node
{
int x;
int y;
}zuo[1005];
int longth(int a,int b)
{
if((zuo[a].x-zuo[b].x)*(zuo[a].x-zuo[b].x)+(zuo[a].y-zuo[b].y)*(zuo[a].y-zuo[b].y)<=d*d)
return 1;
else
return 0;
}
void intt()
{
for(int i=1;i<=n;i++)
f[i]=i;
}
int getf(int v)
{
if(f[v]==v)
return v;
else
{
f[v]=getf(f[v]);
return f[v];
}
}
void marge(int a,int b)
{
int t1,t2;
t1=getf(a);
t2=getf(b);
if(longth(a,b))
{
if(t1!=t2)
f[t1]=t2;
}
return ;
}
int main()
{
int i,j,k,t,t1,t2;
char a;
while(scanf("%d %d",&n,&d)!=EOF)
{
memset(book,0,sizeof(book));
intt();
for(i=1;i<=n;i++)
scanf("%d %d",&zuo[i].x,&zuo[i].y);
while(scanf("%c",&a)!=EOF)
{
if(a=='O')
{
scanf("%d",&t);
book[t]=1;
for(i=1;i<=n;i++)
{
if(book[i]==1)//由于此题不是一次性的把所有的点都用上，因此加了个记录数组
marge(t,i);//将给出的点直接联合（函数中多了一个判断）
}
}
if(a=='S')
{
scanf("%d %d",&t1,&t2);
if(getf(t1)==getf(t2))
printf("SUCCESS\n");
else
printf("FAIL\n");
}
}
}
return 0;
}