hdu4819-Mosaic 二维线段树

Mosaic

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1891    Accepted Submission(s): 831

Problem Description

The God of sheep decides to pixelate some pictures (i.e., change them into pictures with mosaic). Here's how he is gonna make it: for each picture, he divides the picture into n x n cells, where each cell is assigned a color value. Then he chooses a cell, and checks the color values in the L x L region whose center is at this specific cell. Assuming the maximum and minimum color values in the region is A and B respectively, he will replace the color value in the chosen cell with floor((A + B) / 2).

Can you help the God of sheep?

 

Input

The first line contains an integer T (T ≤ 5) indicating the number of test cases. Then T test cases follow.

Each test case begins with an integer n (5 < n < 800). Then the following n rows describe the picture to pixelate, where each row has n integers representing the original color values. The j-th integer in the i-th row is the color value of cell (i, j) of the picture. Color values are nonnegative integers and will not exceed 1,000,000,000 (10^9).

After the description of the picture, there is an integer Q (Q ≤ 100000 (10^5)), indicating the number of mosaics.

Then Q actions follow: the i-th row gives the i-th replacement made by the God of sheep: xi, yi, Li (1 ≤ xi, yi ≤ n, 1 ≤ Li < 10000, Li is odd). This means the God of sheep will change the color value in (xi, yi) (located at row xi and column yi) according to the Li x Li region as described above. For example, an query (2, 3, 3) means changing the color value of the cell at the second row and the third column according to region (1, 2) (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4). Notice that if the region is not entirely inside the picture, only cells that are both in the region and the picture are considered.

Note that the God of sheep will do the replacement one by one in the order given in the input.

 

Output

For each test case, print a line "Case #t:"(without quotes, t means the index of the test case) at the beginning.

For each action, print the new color value of the updated cell.

 

Sample Input

  
  

   
   1
3
1 2 3
4 5 6
7 8 9
5
2 2 1
3 2 3
1 1 3
1 2 3
2 2 3
  
  

 

Sample Output

  
  

   
   Case #1:
5
6
3
4
6
  
  
  
  


  
  
  
  

   
   题意：给你一个n*n的矩阵，给你一个坐标（x，y），询问在坐标附近l的范围内，最大值a和最小值b，将（x，y）的值改为（a+b）/2;
  
  
  
  

   
   

  
  
  
  

   
   解题思路：这是一道关于二维线段树的模板题，一维线段树是解决区间有关的问题，而二维线段树解决的是关于区域的问题，一维线段树是一颗二叉树，而二维线段树是一颗四叉树，非叶子节点有2个或者4个子树，有关二维线段树的具体知识我会稍后写出有关博客，二维线段树专题下面是ac代码：
  
  
  
  

   
   

  
  
  
  

   
   
#include<iostream>
#include<cstdio>
#include <math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#define MOD 10000007
#define maxn 3500
#define LL long long
using namespace std;
int tree_max[maxn][maxn],tree_min[maxn][maxn],ans_min,ans_max,n;
int a[maxn][maxn];
void build_y(int xx,int node,int l,int r,int x,int type)
{
        if(l+1==r){
                if(type)tree_max[xx][node]=tree_min[xx][node]=a[x][l];
                else
                {
                        tree_max[xx][node]=max(tree_max[xx*2][node],tree_max[xx*2+1][node]);
                        tree_min[xx][node]=min(tree_min[xx*2][node],tree_min[xx*2+1][node]);
                }
                return ;
        }
        int mid=(l+r)>>1;
        build_y(xx,node*2,l,mid,x,type);
        build_y(xx,node*2+1,mid,r,x,type);
        tree_max[xx][node]=max(tree_max[xx][node*2],tree_max[xx][node*2+1]);
        tree_min[xx][node]=min(tree_min[xx][node*2],tree_min[xx][node*2+1]);
}
void build_x(int node,int l,int r)
{
        if(l+1==r)
        {
                build_y(node,1,1,n+1,l,1);
                return ;
        }
        int mid=(l+r)>>1;
        build_x(node*2,l,mid);
        build_x(node*2+1,mid,r);
        build_y(node,1,1,n+1,l,0);
}
void query_y(int xx,int node,int l,int r,int ql,int qr)
{
        if(ql<=l&&r<=qr)
        {
                ans_max=max(ans_max,tree_max[xx][node]);
                ans_min=min(ans_min,tree_min[xx][node]);
                return;
        }
        int mid=(l+r)>>1;
        if(ql<mid)query_y(xx,node*2,l,mid,ql,qr);
        if(qr>mid)query_y(xx,node*2+1,mid,r,ql,qr);
}
void query_x(int node,int l,int r,int ql,int qr,int y1,int y2)
{
        if(ql<=l && r<=qr)
        {
                query_y(node,1,1,1+n,y1,y2);return ;
        }
        int mid=(l+r)>>1;
        if(ql<mid)query_x(node*2,l,mid,ql,qr,y1,y2);
        if(qr>mid)query_x(node*2+1,mid,r,ql,qr,y1,y2);
}
void update_y(int xx,int node,int l,int r,int pos,int num,int type){
        if(l+1==r){
                if(type)tree_max[xx][node]=tree_min[xx][node]=num;
                else{
                        tree_max[xx][node]=max(tree_max[xx*2][node],tree_max[xx*2+1][node]);
                        tree_min[xx][node]=min(tree_min[xx*2][node],tree_min[xx*2+1][node]);
                }
                return ;
        }
        int mid=(l+r)>>1;
        if(pos<mid)update_y(xx,node*2,l,mid,pos,num,type);else
        update_y(xx,node*2+1,mid,r,pos,num,type);
        tree_max[xx][node]=max(tree_max[xx][node*2],tree_max[xx][node*2+1]);
        tree_min[xx][node]=min(tree_min[xx][node*2],tree_min[xx][node*2+1]);
}
void update_x(int node,int l,int r,int x,int y,int num)
{
        if(l+1==r)
        {
                update_y(node,1,1,n+1,y,num,1);
                return ;
        }
        int mid=(l+r)>>1;
        if(x<mid)update_x(node*2,l,mid,x,y,num);
        else update_x(node*2+1,mid,r,x,y,num);
        update_y(node,1,1,n+1,y,num,0);
}
int main()
{
        int t,cas=0;
        scanf("%d",&t);
        while(t--)
        {
                memset(tree_max,0,sizeof(tree_max));
                memset(tree_min,0,sizeof(tree_min));
                printf("Case #%d:\n",++cas);
                int q,x,y,l;
                scanf("%d",&n);
                for(int i=1;i<=n;i++)
                        for(int j=1;j<=n;j++)scanf("%d",&a[i][j]);
                build_x(1,1,n+1);
                scanf("%d",&q);
                while(q--)
                {
                        scanf("%d%d%d",&x,&y,&l);l=(l-1)>>1;
                        ans_max=0;ans_min=0x3f3f3f3f;
                        query_x(1,1,n+1,max(x-l,1),min(x+l+1,n+1),max(y-l,1),min(y+l+1,n+1));
                        update_x(1,1,n+1,x,y,(ans_max+ans_min)>>1);
                        printf("%d\n",(ans_max+ans_min)>>1);
                }
        }
        return 0;
}
   
   
题目链接：点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=4819