poj-2236 Wireless Network （並查集）

解題思路：

    並查集的簡單應用，對每次修好的電腦對其它已經修好的電腦遍歷，

    如果距離小於等於最大通信距離就將他們合併。之後判斷2臺電腦是不是一個集合中就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

這道題WA了好幾發，後來發現思路錯了，重新寫了下：

#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));//初始化爲全部不能工作
       init();
        for(i=1;i<=n;++i)
            cin>>map[i][0]>>map[i][1];//輸入computer的座標
        for(i=1;i<=n;++i)
            for(j=1;j<=i;++j)
            {
                //判斷是否兩臺computer的距離是否小於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;
}

然後後看了一下別人寫的代碼（很不爭氣的看了LOL），發現了有趣的東西：

#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)//剛開始發現了，這裏和我自己寫的不一樣，天真的改成自己寫的，結果WA了；

找了一下原因，發現下面初始化不一樣
{
    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])//此時root[i]的絕對值比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));//就是這裏LOL，直接初始化爲-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;
}