hdu 4819 Mosaic 树套树 模板

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?

InputThe 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.��OutputFor 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：
　　

　　　　

看清楚是单点修改。

题解，树套树，外面一层是那几行，里面是行的列区间，nlognlogn的时间复杂度。

 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<iostream>
 #include<cstdio>

 #define N 807
 #define ll long long
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch>''||ch<''){if (ch=='-') f=-;ch=getchar();}
     while(ch<=''&&ch>=''){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }

 int n,m;
 int a[N][N];
 struct Node
 {
     int mi,mx;
 }tr[N<<][N<<];

 void build_sec(int lct,int p,int l,int r,int fla)
 {
     if (l==r)
     {
         if (fla)tr[lct][p].mi=tr[lct][p].mx=a[fla][l];
         else tr[lct][p].mi=min(tr[lct<<][p].mi,tr[lct<<|][p].mi),tr[lct][p].mx=max(tr[lct<<][p].mx,tr[lct<<|][p].mx);
         return;
     }
     int mid=(l+r)>>;
     build_sec(lct,p<<,l,mid,fla),build_sec(lct,p<<|,mid+,r,fla);
     if (fla)
     {
         tr[lct][p].mi=min(tr[lct][p<<].mi,tr[lct][p<<|].mi);
         tr[lct][p].mx=max(tr[lct][p<<].mx,tr[lct][p<<|].mx);
     }
     else
     {
         tr[lct][p].mi=min(tr[lct<<][p].mi,tr[lct<<|][p].mi);
         tr[lct][p].mx=max(tr[lct<<][p].mx,tr[lct<<|][p].mx);
     }
 }
 void build_fir(int p,int l,int r)
 {
     if (l==r)
     {
         build_sec(p,,,n,l);
         return;
     }
     int mid=(l+r)>>;
     build_fir(p<<,l,mid),build_fir(p<<|,mid+,r);
     build_sec(p,,,n,);
 }
 int query_mi2(int lct,int p,int l,int r,int x,int y)
 {
     if (l==x&&y==r) return tr[lct][p].mi;
     int mid=(l+r)>>;
     if (y<=mid) return query_mi2(lct,p<<,l,mid,x,y);
     else if (x>mid) return query_mi2(lct,p<<|,mid+,r,x,y);
     else return min(query_mi2(lct,p<<,l,mid,x,mid),query_mi2(lct,p<<|,mid+,r,mid+,y));
 }
 int query_mi1(int p,int l,int r,int x,int y,int x1,int y1)
 {
     if (l==x&&r==y) return query_mi2(p,,,n,x1,y1);
     int mid=(l+r)>>;
     if (y<=mid) return query_mi1(p<<,l,mid,x,y,x1,y1);
     else if (x>mid) return query_mi1(p<<|,mid+,r,x,y,x1,y1);
     else return min(query_mi1(p<<,l,mid,x,mid,x1,y1),query_mi1(p<<|,mid+,r,mid+,y,x1,y1));
 }
 int query_mx2(int lct,int p,int l,int r,int x,int y)
 {
     if (l==x&&y==r) return tr[lct][p].mx;
     int mid=(l+r)>>;
     if (y<=mid) return query_mx2(lct,p<<,l,mid,x,y);
     else if (x>mid) return query_mx2(lct,p<<|,mid+,r,x,y);
     else return max(query_mx2(lct,p<<,l,mid,x,mid),query_mx2(lct,p<<|,mid+,r,mid+,y));
 }
 int query_mx1(int p,int l,int r,int x,int y,int x1,int y1)
 {
     if (l==x&&r==y) return query_mx2(p,,,n,x1,y1);
     int mid=(l+r)>>;
     if (y<=mid) return query_mx1(p<<,l,mid,x,y,x1,y1);
     else if (x>mid) return query_mx1(p<<|,mid+,r,x,y,x1,y1);
     else return max(query_mx1(p<<,l,mid,x,mid,x1,y1),query_mx1(p<<|,mid+,r,mid+,y,x1,y1));
 }
 void update(int lct,int p,int l,int r,int x)
 {
     tr[lct][p].mi=min(tr[lct<<][p].mi,tr[lct<<|][p].mi);
     tr[lct][p].mx=max(tr[lct<<][p].mx,tr[lct<<|][p].mx);
     if (l==r)return;
     int mid=(l+r)>>;
     if (x<=mid) update(lct,p<<,l,mid,x);
     else update(lct,p<<|,mid+,r,x);
 }
 void modify_sec(int lct,int p,int l,int r,int x,int z)
 {
     if (l==r){tr[lct][p].mi=tr[lct][p].mx=z;return;}
     int mid=(l+r)>>;
     if (x<=mid) modify_sec(lct,p<<,l,mid,x,z);
     else modify_sec(lct,p<<|,mid+,r,x,z);
     tr[lct][p].mi=min(tr[lct][p<<].mi,tr[lct][p<<|].mi);
     tr[lct][p].mx=max(tr[lct][p<<].mx,tr[lct][p<<|].mx);
 }
 void modify_fir(int p,int l,int r,int x,int y,int z)
 {
     if(l==r){modify_sec(p,,,n,y,z);return;}
     int mid=(l+r)>>;
     if (x<=mid)modify_fir(p<<,l,mid,x,y,z);
     else modify_fir(p<<|,mid+,r,x,y,z);
     update(p,,,n,y);
 }
 int main()
 {
     freopen("fzy.in","r",stdin);
     freopen("fzy.out","w",stdout);

     int T=read(),Case=;
     while(T--)
     {
         printf("Case #%d:\n",++Case);
         n=read();
         for (int i=;i<=n;i++)
             for (int j=;j<=n;j++)
                 a[i][j]=read();
         build_fir(,,n);
         m=read();
         for (int i=;i<=m;i++)
         {
             int x=read(),y=read(),z=read()/;
             int x1=max(x-z,),x2=min(x+z,n),y1=max(y-z,),y2=min(y+z,n);
             int ansmi=query_mi1(,,n,x1,x2,y1,y2);
             int ansmx=query_mx1(,,n,x1,x2,y1,y2);
             int ans=(ansmx+ansmi)/;
             printf("%d\n",ans);
             modify_fir(,,n,x,y,ans);
         }
     }
 }