HDU 4819 Mosaic D-section Tree

pid=4819 "> Click to open Link
MosaicTime limit:10000/5000 MS (java/others) Memory limit:102400/102400 K (java/others)
Total submission (s): 657 Accepted Submission (s): 248


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

Source2013 Asia Regional Changchun
give you a n*n matrix and then give you m a request. Each request has x,y,r this three number. Represents a maximum and a minimum value in a matrix with a length of L in the center of Line x, Y, and changes (x, y) to/2 (max + min). A one-dimensional representation of a row in a two-dimensional line segment tree, and one-dimensional representation of the column.

1859ms50772k#include<stdio.h> #include <algorithm> #define M 1007#define EPS 1e-4#define inf    0x3f3f3f3fusing namespace Std;int lx[m],ly[m];int n;struct sub_tree{int left,right,minn,maxx; int mid () {return (left+right) >>1;}};    struct tree{int left,right;    int mid () {return (left+right) >>1;} Sub_tree subtree[4*m];}    Tree[m*4];void build_subtree (int l,int r,int i,int fa) {tree[fa].subtree[i].left=l;    Tree[fa].subtree[i].right=r;    Tree[fa].subtree[i].minn=inf;    Tree[fa].subtree[i].maxx=-inf;    if (l==r) {Ly[l]=i;return;}    int mid= (L+R) >>1;    Build_subtree (L,MID,I&LT;&LT;1,FA); Build_subtree (Mid+1,r, (i<<1) |1,FA);}    void build (int l,int r,int i) {tree[i].left=l;tree[i].right=r;    Build_subtree (1,n,1,i);    if (l==r) {Lx[l]=i;return;}    int mid= (L+R) >>1;    Build (l,mid,i<<1); Build (Mid+1,r, (i<<1) | |);}    void update (int x,int y,int val) {int xx=lx[x];    int yy=ly[y]; tree[xx].subtree[yy].minn=tree[xx].subtree[Yy].maxx=val;            for (int i=xx;i;i>>=1) for (int j=yy;j;j>>=1) {if (I==XX&AMP;&AMP;J==YY) continue; if (j==yy) {tree[i].subtree[j].minn=min (tree[i<<1].subtree[j].minn,tree[(i<<1                ) |1].subtree[j].minn);            Tree[i].subtree[j].maxx=max (tree[i<<1].subtree[j].maxx,tree[(i<<1) |1].subtree[j].maxx); } else {tree[i].subtree[j].minn=min (tree[i].subtree[j<<1].minn,tree[i].subtree                [(j<<1) |1].minn);            Tree[i].subtree[j].maxx=max (tree[i].subtree[j<<1].maxx,tree[i].subtree[(j<<1) |1].maxx); }}}int query_subtree_min (int a1,int a2,int I,int fa) {if (tree[fa].subtree[i].left==a1&&tree[fa].subtree    [I].RIGHT==A2] return tree[fa].subtree[i].minn;    int Mid=tree[fa].subtree[i].mid ();    if (A2<=mid) return query_subtree_min (A1,A2,I&LT;&LT;1,FA); else if (MID&LT;A1) return Query_subtree_min (A1, a2, (i<<1) |1,FA); else return min (query_subtree_min (A1,MID,I&LT;&LT;1,FA), Query_subtree_min (MID+1,A2, (i<<1) |1,FA));} int query_min (int x1,int x2,int y1,int y2,int i) {if (tree[i].left==x1&&tree[i].right==x2) return query_subtree_    Min (y1,y2,1,i);    int Mid=tree[i].mid ();    if (X2<=mid) return query_min (x1,x2,y1,y2,i<<1);    else if (mid<x1) return Query_min (X1,x2,y1,y2, (i<<1) | | |); else return min (query_min (x1,mid,y1,y2,i<<1), Query_min (Mid+1,x2,y1,y2, (i<<1) | |));} int Query_subtree_max (int a1,int a2,int I,int fa) {if (Tree[fa].subtree[i].left==a1&&tree[fa].subtree[i].    RIGHT==A2) return Tree[fa].subtree[i].maxx;    int Mid=tree[fa].subtree[i].mid ();    if (A2<=mid) return Query_subtree_max (A1,A2,I&LT;&LT;1,FA);    else if (MID&LT;A1) return Query_subtree_max (A1,A2, (i<<1) |1,FA); else return Max (Query_subtree_max (A1,MID,I&LT;&LT;1,FA), Query_subtree_max (MID+1,A2, (i<<1) |1,FA));} int Query_max (int x1,int x2,int y1,int y2,inT i) {if (tree[i].left==x1&&tree[i].right==x2) return Query_subtree_max (y1,y2,1,i);    int Mid=tree[i].mid ();    if (X2<=mid) return Query_max (x1,x2,y1,y2,i<<1);    else if (mid<x1) return Query_max (X1,x2,y1,y2, (i<<1) | | |); else return Max (Query_max (x1,mid,y1,y2,i<<1), Query_max (Mid+1,x2,y1,y2, (i<<1) | |));}    int main () {int t,cas=1;    scanf ("%d", &t);       while (t--) {int m,a,x,y,r;       scanf ("%d", &n);       Build (1,n,1);                for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) {scanf ("%d", &a);            Update (I,J,A);        } printf ("Case #%d:\n", cas++);        scanf ("%d", &m);            while (m--) {scanf ("%d%d%d", &x,&y,&r);            int X1=max (1,X-R/2);            int x2=min (N,X+R/2);            int Y1=max (1,Y-R/2);            int y2=min (N,Y+R/2);            int Maxx=query_max (x1,x2,y1,y2,1);     int minn=query_min (x1,x2,y1,y2,1);       int ans= (MINN+MAXX) >>1;            printf ("%d\n", ans);        Update (X,y,ans); }} return 0;}

Copyright notice: This article Bo Master original articles, blogs, without consent may not be reproduced.

HDU 4819 Mosaic D-section Tree