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  • ARTWORK

    Description

    A template for an artwork is a white grid of n × m squares. The artwork will be created by painting q horizontal and vertical black strokes. A stroke starts from square (x1, y1), ends at square (x2,y2)(x1=x2(x2,y2)(x1=x2 or y1=y2)y1=y2) and changes the color of all squares (x, y) to black where x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2. The beauty of an artwork is the number of regions in the grid. Each region consists of one or more white squares that are connected to each other using a path of white squares in the grid, walking horizontally or vertically but not diagonally. The initial beauty of the artwork is 1. Your task is to calculate the beauty after each new stroke. Figure A.1 illustrates how the beauty of the artwork varies in Sample Input 1.

    Input

    The first line of input contains three integers n, m and q (1 ≤ n, m ≤ 1000, 1 ≤ q ≤ 104). Then follow q lines that describe the strokes. Each line consists of four integersx1,y1,x2x1,y1,x2 and y2(1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ m). Either x1 = x2 or y1=y2y1=y2(or both).

    Output

    For each of the q strokes, output a line containing the beauty of the artwork after the stroke.

    Sample Input

    4 6 5
    2 2 2 6
    1 3 4 3
    2 5 3 5
    4 6 4 6
    1 6 4 6

    Sample Output

    1
    3
    3
    4
    3

    代码为:

    #include<iostream>
    #include<cstdio> 
    #include<cstring>
    using namespace std;
    
    
    int n,m,q;
    
    
    struct node{
        int x1,y1,xx,yy;
    } que[10005];
    
    
    int num[1000005];
    int f[1000005];
    
    
    int find(int x)
    {
        return f[x]==x?x:f[x]=find(f[x]);
    }
    
    
    int cnt;
    void merge(int x,int y)
    {
        int fx=find(x);
        int fy=find(y);
        if(fx==fy) 
    return;
        cnt--; 
    f[fx]=fy;
    }
    
    
    int Get_id(int x,int y)
    {
         return x*m+y;
    }
    
    
    int ans[10005];
    
    
    int dfsx[4]={0,0,1,-1};
    int dfsy[4]={1,-1,0,0};
    
    
    bool in(int x,int y)
    {
        return x>=0&&y>=0&&x<n&&y<m;
    }
    void work(int x,int y)
    {
        for(int k=0;k<4;k++)
    {
            int tx=x+dfsx[k];
            int ty=y+dfsy[k];
            if(!in(tx,ty)||num[Get_id(tx,ty)]) 
      continue;
            merge(Get_id(tx,ty),Get_id(x,y));
        }
    }
    int main()
    {
        cin>>m>>n>>q;
        cnt=n*m;
        
        for(int i=0;i<n*m;i++) 
    f[i]=i,num[i]=0;
        for(int i=0;i<q;i++)
    {
            cin>>que[i].y1>>que[i].x1>>que[i].yy>>que[i].xx;
            que[i].x1--; 
    que[i].y1--;
            que[i].xx--; 
    que[i].yy--;
            for(int x=que[i].x1;x<=que[i].xx;x++)
            {
    for(int y=que[i].y1;y<=que[i].yy;y++)
    {
               if(num[Get_id(x,y)]==0) cnt--;
                num[Get_id(x,y)]++;
            }
            }
        }
        for(int i=0;i<n;i++)
    {
            for(int j=0;j<m;j++)
       {
                if(num[Get_id(i,j)]) 
      continue;
                work(i,j);
            }
        }
        for(int i=q-1;i>=0;i--)
    {
            ans[i]=cnt;
            for(int x=que[i].x1;x<=que[i].xx;x++)
            for(int y=que[i].y1;y<=que[i].yy;y++)
    {
                num[Get_id(x,y)]--;
                if(num[Get_id(x,y)]==0) 
    {
                    cnt++;
                    work(x,y);
                }
            }
        }
    
    
        for(int i=0;i<q;i++)
            cout<<ans[i]<<endl;
    
    
       return 0;
    }
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  • 原文地址:https://www.cnblogs.com/csushl/p/9386578.html
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