hdoj 48119 Mosaic (二维线段树 单点更新)

Mosaic

点击打开链接

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1016    Accepted Submission(s): 449

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

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

 

Sample Output

Case #1:56346

 

题意：

给出一个图片，分成N×N个单元格，有M次操作，每次操作将(x,y)的值变为以(x,y) 为中心L(L为奇数)为边长的区域内的最小值和最大值的均值(floor((maximum+minimum)/2)),并输出该值。

分析：

明显的二维线段树的单点更新和区间查询，维护最值。

更新肯定是先在二维内找到叶子节点的那棵线段树，然后再在这棵树上更新，这部分很简单，就想成一维的写。但是这毕竟是二维线段树，二维的部分也需要维护，当然这部分比较麻烦，我们先想一想一维的：一维的节点维护的是值，我们只要根据它的左右儿子的信息维护(pushup)就行。二维的节点维护的是树(一维的线段树),类比一下，我们也需要根据它的左右儿子来维护，但是我们并不能简单地通过pushup来维护它的信息(因为一个节点对应的是一棵一维线段树，而不是一个线段)，我们要进入这棵一维线段树中去维护各个点(线段)的信息。

维护好过后查询就方便了，简单地降维即可。

这是我看着匡冰大神的博客写的，第一次写这玩意儿，自己写老是错，求大神指点

我写的代码

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>using namespace std;#define MAX  1010#define inf 0x7fffffffstruct node{int L;int R;int vmax;int vmin;};struct node1{int hl;int hr;node subt[4*MAX];}t[4*MAX];void build_subt(int v,int rt,int l,int r){ t[v].subt[rt].L=l; t[v].subt[rt].R=r; t[v].subt[rt].vmax=-inf; t[v].subt[rt].vmin=inf; if(l==r)   return ;  int mid=(l+r)>>1;  build_subt(v,rt<<1,l,mid);  build_subt(v,rt<<1|1,mid+1,r);} void build(int v,int l,int r,int ll,int rr){t[v].hl=l;t[v].hr=r;build_subt(v,1,ll,rr);if(l==r)return ;int mid=(l+r)>>1;build(v<<1,l,mid,ll,rr);build(v<<1|1,mid+1,r,ll,rr);}void update_subt(int v,int rt,int pos,int x){t[v].subt[rt].vmax=max(x,t[v].subt[rt].vmax); t[v].subt[rt].vmin=min(x,t[v].subt[rt].vmin); if(t[v].subt[rt].L==t[v].subt[rt].R)return ;int mid=(t[v].subt[rt].L+t[v].subt[rt].R)>>1;if(pos<=mid)update_subt(v,rt<<1,pos,x);elseupdate_subt(v,rt<<1|1,pos,x);t[v].subt[rt].vmax=max(t[v].subt[rt<<1].vmax,t[v].subt[rt<<1|1].vmax);t[v].subt[rt].vmin=min(t[v].subt[rt<<1].vmin,t[v].subt[rt<<1|1].vmin);}void update(int v,int cur,int x,int y){update_subt(v,1,x,y);if(t[v].hl==t[v].hr)return ;int mid=(t[v].hl+t[v].hr)>>1;if(cur<=mid)update(v<<1,cur,x,y);elseupdate(v<<1|1,cur,x,y);}int query_subtmax(int v,int rt,int l,int r){if(t[v].subt[rt].L==l&&t[v].subt[rt].R==r)return t[v].subt[rt].vmax;int mid=(t[v].subt[rt].L+t[v].subt[rt].R)>>1;if(r<=mid) return query_subtmax(v,rt<<1,l,r);else if(l>mid)return query_subtmax(v,rt<<1|1,l,r);elsereturn max(query_subtmax(v,rt<<1,l,mid),query_subtmax(v,rt<<1|1,mid+1,r)); }int querymax(int v,int l,int r,int ll,int rr){if(t[v].hl==l&&t[v].hr==r)return query_subtmax(v,1,ll,rr);int mid=(t[v].hl+t[v].hr)>>1;if(r<=mid)return querymax(v<<1,l,r,ll,rr);else if(l>mid)return querymax(v<<1|1,l,r,ll,rr);else return max(querymax(v<<1,l,mid,ll,rr),querymax(v<<1|1,mid+1,r,ll,rr));}int query_subtmin(int v,int rt,int l,int r){if(t[v].subt[rt].L==l&&t[v].subt[rt].R==r)return t[v].subt[rt].vmin;int mid=(t[v].subt[rt].L+t[v].subt[rt].R)>>1;if(r<=mid) return query_subtmin(v,rt<<1,l,r);else if(l>mid)return query_subtmin(v,rt<<1|1,l,r);elsereturn min(query_subtmin(v,rt<<1,l,mid),query_subtmin(v,rt<<1|1,mid+1,r)); }int querymin(int v,int l,int r,int ll,int rr){if(t[v].hl==l&&t[v].hr==r)return query_subtmin(v,1,ll,rr);int mid=(t[v].hl+t[v].hr)>>1;if(r<=mid)return querymin(v<<1,l,r,ll,rr);else if(l>mid)return querymin(v<<1|1,l,r,ll,rr);else return min(querymin(v<<1,l,mid,ll,rr),querymin(v<<1|1,mid+1,r,ll,rr));}int main(){int T,ca=1;scanf("%d",&T);while(T--){  int n;  scanf("%d",&n);  build(1,1,n,1,n);    for(int i=1;i<=n;i++)   for(int j=1;j<=n;j++)   {   int v;   scanf("%d",&v);update(1,v,i,j);   }   int Q;   scanf("%d",&Q);   printf("Case #%d:\n",ca++);   while(Q--)   {   int xi,yi,Li;   scanf("%d%d%d",&xi,&yi,&Li);         int x1 = max(xi - Li/2,1);         int x2 = min(xi + Li/2,n);         int y1 = max(yi - Li/2,1);         int y2 = min(yi + Li/2,n);         if(x1>x2)         swap(x1,x2);         if(y1>y2)         swap(y1,y2); int Max = querymax(1,x1,x2,y1,y2);         int Min = querymin(1,x1,x2,y1,y2);         int t = (Max+Min)/2;         printf("%d\n",t);         update(1,t,xi,yi);   }}return 0;}

匡冰大神版本，大神就是大神:点击打开链接

#include<cstdio>#include<queue>#include<cstring>#include<cstdlib>#include<algorithm>using namespace std;#define inf 0x7ffffff#define MAXN 1010int n;struct nodey{int ly;int ry;int vmax;int vmin;};int locate_y[MAXN],locate_x[MAXN];struct nodex{int lx,rx;nodey sty[MAXN*4];void build(int l,int r,int v){sty[v].ly=l;sty[v].ry=r;sty[v].vmax=-inf;sty[v].vmin=inf;if(l==r){ locate_y[l]=v; return ;}    int mid=(l+r)>>1;    build(l,mid,v<<1);    build(mid+1,r,v<<1|1);}      int querymin(int l,int r,int v)   {   if(sty[v].ly==l&&sty[v].ry==r)   return sty[v].vmin;      int mid=(sty[v].ly+sty[v].ry)>>1;   if(r<=mid)   return querymin(l,r,v<<1);   else if(l>mid)   return querymin(l,r,v<<1|1);   else   return min(querymin(l,mid,v<<1),querymin(mid+1,r,v<<1|1));   }      int querymax(int l,int r,int v)   {   if(sty[v].ly==l&&sty[v].ry==r)   return sty[v].vmax;      int mid=(sty[v].ly+sty[v].ry)>>1;   if(r<=mid)   return querymax(l,r,v<<1);   else if(l>mid)   return querymax(l,r,v<<1|1);   else   return max(querymax(l,mid,v<<1),querymax(mid+1,r,v<<1|1));   }}stx[MAXN*4];void build(int l,int r,int v){stx[v].lx=l;stx[v].rx=r;stx[v].build(1,n,1);if(l==r){locate_x[l]=v;return ;}int mid=(l+r)>>1;build(l,mid,v<<1);build(mid+1,r,v<<1|1);}void update(int x,int y,int va){int tx=locate_x[x];int ty=locate_y[y];stx[tx].sty[ty].vmin=stx[tx].sty[ty].vmax=va;for(int i=tx;i;i>>=1)  for(int j=ty;j;j>>=1)  {  if(i==tx&&j==ty)  continue;  if(j==ty)  {    stx[i].sty[j].vmin=min(stx[i<<1].sty[j].vmin,stx[i<<1|1].sty[j].vmin);  stx[i].sty[j].vmax=max(stx[i<<1].sty[j].vmax,stx[i<<1|1].sty[j].vmax);}else{    stx[i].sty[j].vmin=min(stx[i].sty[j<<1].vmin,stx[i].sty[j<<1|1].vmin);  stx[i].sty[j].vmax=max(stx[i].sty[j<<1].vmax,stx[i].sty[j<<1|1].vmax);}  }}int querymin(int x1,int x2,int y1,int y2,int v){if(stx[v].lx==x1&&stx[v].rx==x2)return stx[v].querymin(y1,y2,1);int mid=(stx[v].lx+stx[v].rx)>>1;if(x2<=mid)return querymin(x1,x2,y1,y2,v<<1);else if(x1>mid)return querymin(x1,x2,y1,y2,v<<1|1);elsereturn min(querymin(x1,mid,y1,y2,v<<1),querymin(mid+1,x2,y1,y2,v<<1|1));}int querymax(int x1,int x2,int y1,int y2,int v){if(stx[v].lx==x1&&stx[v].rx==x2)return stx[v].querymax(y1,y2,1);int mid=(stx[v].lx+stx[v].rx)>>1;if(x2<=mid)return querymax(x1,x2,y1,y2,v<<1);else if(x1>mid)return querymax(x1,x2,y1,y2,v<<1|1);elsereturn max(querymax(x1,mid,y1,y2,v<<1),querymax(mid+1,x2,y1,y2,v<<1|1));}int main(){ int T,ca=1; scanf("%d",&T); while(T--) { scanf("%d",&n); build(1,n,1); for(int i=1;i<=n;i++)  for(int j=1;j<=n;j++)  {    int v;   scanf("%d",&v);   update(i,j,v); } int Q; scanf("%d",&Q); printf("Case #%d:\n",ca++); while(Q--) { int xi,yi,Li; scanf("%d%d%d",&xi,&yi,&Li); int x1=max(xi-Li/2,1); int x2=min(xi+Li/2,n); int y1=max(yi-Li/2,1); int y2=min(yi+Li/2,n); int ret_max=querymax(x1,x2,y1,y2,1); int ret_min=querymin(x1,x2,y1,y2,1); int ans=(ret_max+ret_min)>>1; printf("%d\n",ans); update(xi,yi,ans); } } return 0;}