• hihocoder Counting Islands II(并查集)


    Counting Islands II

    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.  

    输入

    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)

    输出

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

    样例输入
    3  
    0 0   
    1 1   
    1 0   
    样例输出
    1  
    2  
    1
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <list>
    #include <ext/rope>
    #define rep(i,m,n) for(i=m;i<=n;i++)
    #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
    #define vi vector<int>
    #define pii pair<int,int>
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define ll long long
    #define pi acos(-1.0)
    const int maxn=1e5+10;
    using namespace std;
    using namespace __gnu_cxx;
    const int N = 1100005;
    int pre[N],ans[1005][1005];
    int d[4][2]= {0,1,1,0,0,-1,-1,0};
    int n,cnt=0;
    set<int>sa;
    int Find(int x)
    {
        if(pre[x] != x)
            pre[x] = Find(pre[x]);
        return pre[x];
    }
    void init()
    {
        for(int i=0; i<N; i++)pre[i]=i;
    }
    int main()
    {
        init();
        int X,Y;
        int g=0;
        cin>>n;
        for(int i=0; i<n; i++)
        {
            g++;
            cin>>X>>Y;
            ans[X][Y]=1;
            for(int j=0; j<4; j++)
            {
                int xx=X+d[j][0];
                int yy=Y+d[j][1];
                if(xx>=0&&xx<1000&&yy>=0&&yy<1000)
                {
                    if(ans[xx][yy]==1)
                    {
                        int aa=Find(xx*1000+yy),bb=Find(X*1000+Y);
                        if(aa!=bb)pre[aa]=bb,g--;
                    }
                }
            }
            cout<<g<<endl;
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/jianrenfang/p/5711759.html
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