Hdu-4819-mosaic (two-dimensional line segment tree)

Problem DescriptionThe God of sheep decides to pixelate some pictures (i.e., change them into pictures with mosaic). Here's how he's gonna make it:for each picture, he divides the picture to n x n cells, where each cell is assigned a C Olor value. Then he chooses a cell, and checks the color values of the L x L region whose center are at this specific cell. Assuming the maximum and minimum color values in the region is A and B respectively, he'll replace the color value in th e 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 is 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 was the color value of the cell (I, j) of the picture. The Color values are nonnegative integers and would 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≤l I < 10000, Li is odd). This means the God of sheep would 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, a query (2, 3, 3) means changing the color value of the cell at the second row and the third column Accordin G to region (1, 2) (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4). Notice that if the A is not a entirely inside the picture, but cells is both Onsidered.

Note that the God of sheep would do the replacement one by one in the order given in the Input.
Outputfor, 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

131 2 34 5 67 8 952 2 13 2 31 1 31 2 32 2 3

Sample Output

Case #1:56346

Source

Field=problem&key=2013%20asia%20regional%20changchun%20&source=1&searchmode=source ">2013 Asia Regional Changchun

Idea: two-dimensional line segment tree, single point update.

#include <stdio.h> #define INF 1000000001#define min (A, b) (a<b?) A:B) #define MAX (A, b) (a>b? A:B) int mx[3200][3200],mn[3200][3200],n,minnum,maxnum;void buildy (int x,int idx,int s,int e,bool flag)// A flag of 1 indicates that the first dimension is a leaf node.            0 indicates that the first dimension is not a leaf node {if (s==e) {if (flag) {scanf ("%d", &mn[x][idx]);        MX[X][IDX]=MN[X][IDX];        }} else {int mid= (s+e) >>1;        Buildy (X,idx<<1,s,mid,flag);        Buildy (X,idx<<1|1,mid+1,e,flag);        Mx[x][idx]=max (mx[x][idx<<1],mx[x][idx<<1|1]);    Mn[x][idx]=min (mn[x][idx<<1],mn[x][idx<<1|1]);        } if (!flag)//The first dimension is not a leaf node to update the first dimension {Mx[x][idx]=max (Mx[x<<1][idx],mx[x<<1|1][idx]);    Mn[x][idx]=min (Mn[x<<1][idx],mn[x<<1|1][idx]);        }}void Build (int idx,int s,int e) {if (s!=e) {int mid= (s+e) >>1;        Build (Idx<<1,s,mid);        Build (Idx<<1|1,mid+1,e);    Buildy (idx,1,1,n,0); } ElSe buildy (idx,1,1,n,1);}    void Updatey (int x,int idx,int s,int e,int pos,int val,bool flag) {if (s==e) Mn[x][idx]=mx[x][idx]=val;        else {int mid= (s+e) >>1;        if (pos<=mid) Updatey (X,idx<<1,s,mid,pos,val,flag);        else Updatey (X,idx<<1|1,mid+1,e,pos,val,flag);        Mx[x][idx]=max (mx[x][idx<<1],mx[x][idx<<1|1]);    Mn[x][idx]=min (mn[x][idx<<1],mn[x][idx<<1|1]);        } if (!flag)//The first dimension is not a leaf node to update the first dimension {Mx[x][idx]=max (Mx[x<<1][idx],mx[x<<1|1][idx]);    Mn[x][idx]=min (Mn[x<<1][idx],mn[x<<1|1][idx]);    }}void Update (int idx,int s,int e,int pos,int posy,int val) {if (s==e) Updatey (idx,1,1,n,posy,val,1);        else {int mid= (s+e) >>1;        if (pos<=mid) update (idx<<1,s,mid,pos,posy,val);        else update (idx<<1|1,mid+1,e,pos,posy,val);    Updatey (idx,1,1,n,posy,val,0); }}void queryy (int x,int idx,int s,int e,int l,int R) {if (s==l && e==r) {minnum=min (minnum,mn[x][idx]);    Maxnum=max (Maxnum,mx[x][idx]);        } else {int mid= (s+e) >>1;        if (r<=mid) queryy (x,idx<<1,s,mid,l,r);        else if (l>mid) queryy (x,idx<<1|1,mid+1,e,l,r);            else {queryy (x,idx<<1,s,mid,l,mid);        Queryy (X,IDX&LT;&LT;1|1,MID+1,E,MID+1,R);    }}}void query (int idx,int s,int e,int l,int r,int ly,int ry) {if (s==l && e==r) queryy (Idx,1,1,n,ly,ry);        else {int mid= (s+e) >>1;        if (r<=mid) query (Idx<<1,s,mid,l,r,ly,ry);        else if (l>mid) query (Idx<<1|1,mid+1,e,l,r,ly,ry);            else {query (idx<<1,s,mid,l,mid,ly,ry);        Query (Idx<<1|1,mid+1,e,mid+1,r,ly,ry);    }}}int Main () {int t,i,q,a,b,c,l,r,ly,ry,t;    scanf ("%d", &t);        for (i=1;i<=t;i++) {scanf ("%d", &n);        Build (1,1,n);        scanf ("%d", &q); printf ("Case #%d:\n ", i);            while (q--) {scanf ("%d%d%d", &a,&b,&c);            L=max (a-c/2,1);            R=min (A+c/2,n);            Ly=max (b-c/2,1);            Ry=min (B+c/2,n);            Minnum=inf;            Maxnum=-inf;            Query (1,1,n,l,r,ly,ry);            t= (Minnum+maxnum)/2;            printf ("%d\n", t);        Update (1,1,N,A,B,T); }    }}

Hdu-4819-mosaic (two-dimensional line segment tree)