poj2236 (并查集)

Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 38915		Accepted: 16135

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

分析：并查集。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int x[1010],y[1010];
int dis[1004][1004];
int pre[1010],vis[1010];
int N,D;
void init()
{
    for(int i=0;i<=N;i++)
        pre[i]=i;
}

int find(int x)//查找点x所在树的根 
{
    int root=x;
    while(pre[root]!=root)
        root=pre[root];
    int i=x,k;
    while(k!=root)//路径压缩,让x到根经过的所有节点直接指向root
    {
        k=pre[i];
        pre[i]=root;
        i=k;
    }
    return root;
}

void unit(int x,int y)
{
    int rootx=find(x),rooty=find(y);
    if(rootx!=rooty)
    {
        //if(rootx<rooty)
    //        pre[rooty]=pre[rootx];
        //else 
            pre[rootx]=pre[rooty];
    }
}

void Opre(int p)//修复电脑p
{
    for(int i=1;i<=N;i++)
    if(vis[i]&&dis[p][i]<=D&&i!=p)//找孤立的点,root也算 
    {
        unit(i,p);
    }
}

int main()
{
    scanf("%d%d",&N,&D);
    init();
    for(int i=1;i<=N;i++)
    {
        scanf("%d%d",&x[i],&y[i]);    
    }
    D=D*D;
    for(int i=1;i<N;i++)
    for(int j=i+1;j<=N;j++)//两点间距离 
        dis[j][i]=dis[i][j]=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
    char s[10]; 
    while(scanf("%s",s)!=EOF)
    {
        if(s[0]=='O')
        {
            int p;
            scanf("%d",&p);//修复电脑p
            vis[p]=1;
            Opre(p);
            //for(int i=1;i<=N;i++)
            //printf("pre[%d]=%d
",i,pre[i]);
        }
        else
        {
            int p,q;
            scanf("%d%d",&p,&q);//尝试联系电脑p和q 
            if(find(p)==find(q))
                printf("SUCCESS
");
            else printf("FAIL
");
        }
        
    }
    
    return 0;
}