HDU 4819 Mosaic

Question:

Each grid in a rectangle has a value. Now there are Q operations. Each operation provides coordinates (x, y) and length. l each operation outputs (x, y) the Center side length is half of the sum of the maximum and minimum values in the L rectangle and the value is updated to the (x, y) coordinate.

Ideas:

The maximum and minimum values of interval query single point update is obviously a feature of the Line Segment tree, but here it is a two-dimensional line segment tree.

My understanding of the two-dimensional line segment tree: (personal understanding is not necessarily correct)

Initialization effort is equivalent to N * n single-point updates

The Thinking of tree-setting operations first finds the interval that satisfies the first digit and corresponds to several trees, and then operates in these trees (I .e., the second dimension ).

Building and querying are relatively simple, just separate the two dimensions mentioned above.

Updating is troublesome if you want to update (x, y) you must first find X in the first dimension, then assign values to Y in the second dimension corresponding to X, and then perform the Up Operation (this indicates that the interval (x, x) in the first dimension is updated) the last update in the first dimension is to use (x, y) to update all the first-dimensional intervals containing X after Update (it is hard to say that the Code is clearer)

PS:

In the beginning, Changchun had such problems... It's not until one year later... It's been a full afternoon !!

I have nothing to say to myself...

Code:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define inf 2147483647#define N 805#define L(x) (x<<1)#define R(x) ((x<<1)|1)#define Mid(x,y) ((x+y)>>1)struct nodey{    int l,r,minval,maxval;};struct nodex{    int l,r;    nodey y[N*4];}x[N*4];int n,q,lx,ly,ll;void inity(int l,int r,int i,int j){    x[j].y[i].l=l; x[j].y[i].r=r; x[j].y[i].minval=inf; x[j].y[i].maxval=-inf;    if(l==r) return ;    inity(l,Mid(l,r),L(i),j);    inity(Mid(l,r)+1,r,R(i),j);}void initx(int l,int r,int i){    x[i].l=l; x[i].r=r;    inity(1,n,1,i);    if(l==r) return ;    initx(l,Mid(l,r),L(i));    initx(Mid(l,r)+1,r,R(i));}int maxans,minans;void queryy(int l,int r,int i,int j){    if(x[j].y[i].l==l&&x[j].y[i].r==r)    {        maxans=max(maxans,x[j].y[i].maxval);        minans=min(minans,x[j].y[i].minval);        return ;    }    int mid=Mid(x[j].y[i].l,x[j].y[i].r);    if(r<=mid) queryy(l,r,L(i),j);    else if(l>mid) queryy(l,r,R(i),j);    else    {        queryy(l,mid,L(i),j);        queryy(mid+1,r,R(i),j);    }}void queryx(int l,int r,int i){    if(x[i].l==l&&x[i].r==r)    {        queryy(max(1,ly-ll/2),min(n,ly+ll/2),1,i);        return ;    }    int mid=Mid(x[i].l,x[i].r);    if(r<=mid) queryx(l,r,L(i));    else if(l>mid) queryx(l,r,R(i));    else    {        queryx(l,mid,L(i));        queryx(mid+1,r,R(i));    }}void updatey(int l,int i,int j,int key){    if(x[j].y[i].l==x[j].y[i].r)    {        if(x[j].l==x[j].r)        {            x[j].y[i].minval=key;            x[j].y[i].maxval=key;        }        else        {            x[j].y[i].minval=min(x[L(j)].y[i].minval,x[R(j)].y[i].minval);            x[j].y[i].maxval=max(x[L(j)].y[i].maxval,x[R(j)].y[i].maxval);        }        return ;    }    int mid=Mid(x[j].y[i].l,x[j].y[i].r);    if(l<=mid) updatey(l,L(i),j,key);    else updatey(l,R(i),j,key);    x[j].y[i].minval=min(x[j].y[L(i)].minval,x[j].y[R(i)].minval);    x[j].y[i].maxval=max(x[j].y[L(i)].maxval,x[j].y[R(i)].maxval);}void updatex(int l,int i,int key){    if(x[i].l==x[i].r)    {        updatey(ly,1,i,key);        return ;    }    int mid=Mid(x[i].l,x[i].r);    if(l<=mid) updatex(l,L(i),key);    else updatex(l,R(i),key);    updatey(ly,1,i,key);}int main(){    int t,i,j,k,tmp;    scanf("%d",&t);    for(k=1;k<=t;k++)    {        scanf("%d",&n);        initx(1,n,1);        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                scanf("%d",&tmp);                ly=j;                updatex(i,1,tmp);            }        }        scanf("%d",&q);        printf("Case #%d:\n",k);        while(q--)        {            scanf("%d%d%d",&lx,&ly,&ll);            minans=inf; maxans=-inf;            queryx(max(1,lx-ll/2),min(n,lx+ll/2),1);            tmp=Mid(maxans,minans);            printf("%d\n",tmp);            updatex(lx,1,tmp);        }    }    return 0;}