Counting Islands II_太閣競賽C

題目3 : Counting Islands II
時間限制:10000ms
單點時限:1000ms
內存限制:256MB
描述
Country H is going to carry out a huge artificial islands project. The project region is divided into a 1000x1000 grid. The whole project will last for N weeks. Each week one unit area of sea will be filled with land.


As a result, new islands (an island consists of all connected land in 4 -- up, down, left and right -- directions) emerges in this region. Suppose the coordinates of the filled units are (0, 0), (1, 1), (1, 0). Then after the first week there is one island:  


#...
....
....
....
After the second week there are two islands:  


#...
.#..
....
....
After the three week the two previous islands are connected by the newly filled land and thus merge into one bigger island:


#...
##..
....
....
Your task is track the number of islands after each week's land filling.  


輸入
The first line contains an integer N denoting the number of weeks. (1 ≤ N ≤ 100000)  


Each of the following N lines contains two integer x and y denoting the coordinates of the filled area.  (0 ≤ x, y < 1000)


輸出
For each week output the number of islands after that week's land filling.


樣例輸入
3  
0 0   
1 1   
1 0   
樣例輸出
1  
2  

1  

【我的程序】

#include <iostream>
using namespace std;

long int n, num=0, a[1000000]={0}, aSize[1000000]={0};
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

long int findID(long int x)
{
    while (a[x]!=x) x=a[x]=a[a[x]];
    return x;
}

void unionID(long int x, long int y)
{
    x=findID(x); y=findID(y);
    if (x==y) return;
    if (aSize[x]<aSize[y]) {a[x]=y; aSize[y]+=aSize[x];}
    else {a[y]=x; aSize[x]+=aSize[y];}
    num--;
}

int check(int x, int y)
{
    if (x>=0 && x<1000 && y>=0 && y<1000 && aSize[x*1000+y]>0) return 1; else return 0;
}

int main()
{
    long int i,p,q;
    cin>>n;

    for (i=0;i<n;i++)
    {
        cin>>p >>q;
        a[p*1000+q]=p*1000+q;
        aSize[p*1000+q]=1;
        num++;

        for (int j=0;j<4;j++)
            if (check(p+dx[j],q+dy[j])) unionID(p*1000+q,(p+dx[j])*1000+q+dy[j]);

        cout<<num <<endl;
    }

    return 0;
}