poj-2236 Wireless Network （并查集）

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

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

这道题WA了好几发，后来发现思路错了，重新写了下：

#include<iostream>
#include<cstring>
#include<cmath>
int n,a,b,d;
char s;
int map[1002][2],father[1002];
bool dis[1002][1002],work[1002];
void init()
{
    for(int i=1;i<=n;i++)
        father[i]=i;
}
int cal(int x1,int y1,int x2,int y2)
{
    return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}

int find_father(int i)
{
    if(father[i]==i)
        return i;
    else
        father[i]=find_father(father[i]);
        return father[i];
}

void merge(int i,int j)
{
   int t1,t2;
   t1=find_father(i);
   t2=find_father(j);
   if(t1!=t2)
   {
       father[t2]=t1;
   }
   return ;
}

int main()
{
    using namespace std;
    int i,j;
    while(cin>>n>>d)
    {
        memset(work,false,sizeof(work));//初始化为全部不能工作
       init();
        for(i=1;i<=n;++i)
            cin>>map[i][0]>>map[i][1];//输入computer的座标
        for(i=1;i<=n;++i)
            for(j=1;j<=i;++j)
            {
                //判断是否两台computer的距离是否小于d
                int t=cal(map[i][0],map[i][1],map[j][0],map[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_father(a)==find_father(b))
                    cout<<"SUCCESS"<<endl;
                else
                    cout<<"FAIL"<<endl;
            }
        }
    }
    return 0;
}

然后后看了一下别人写的代码（很不争气的看了LOL），发现了有趣的东西：

#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)//刚开始发现了，这里和我自己写的不一样，天真的改成自己写的，结果WA了；

找了一下原因，发现下面初始化不一样
{
    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));//就是这里LOL，直接初始化为-1了。。。
        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;
}