HDU4819 Mosaic（二维线段树）

题意:

给定一个n*n的矩阵,每次给定一个子矩阵区域(x,y,l),求出该区域内的最大值(A)和最小值(B),输出(A+B)/2,并用这个值更新矩阵[x,y]的值

分析：

二维线段树的水题了。

对于二维的矩阵，需要查询一个区域的最大和最小值。

修改单个点的值。

 

二维线段树直接搞，主要是修改的时候，更新操作要往两个方向进行。

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const int INF = 0x3f3f3f3f;const int MAXN = 1010;struct Nodey{    int l,r;    int Max,Min;};int locy[MAXN],locx[MAXN];struct Nodex{    int l,r;    Nodey sty[MAXN*4];    void build(int i,int _l,int _r)    {        sty[i].l = _l;        sty[i].r = _r;        sty[i].Max = -INF;        sty[i].Min = INF;        if(_l == _r)        {            locy[_l] = i;            return;        }        int mid = (_l + _r)/2;        build(i<<1,_l,mid);        build((i<<1)|1,mid+1,_r);    }    int queryMin(int i,int _l,int _r)    {        if(sty[i].l == _l && sty[i].r == _r)            return sty[i].Min;        int mid = (sty[i].l + sty[i].r)/2;        if(_r <= mid)return queryMin(i<<1,_l,_r);        else if(_l > mid)return queryMin((i<<1)|1,_l,_r);        else return min(queryMin(i<<1,_l,mid),queryMin((i<<1)|1,mid+1,_r));    }    int queryMax(int i,int _l,int _r)    {        if(sty[i].l == _l && sty[i].r == _r)            return sty[i].Max;        int mid = (sty[i].l + sty[i].r)/2;        if(_r <= mid)return queryMax(i<<1,_l,_r);        else if(_l > mid)return queryMax((i<<1)|1,_l,_r);        else return max(queryMax(i<<1,_l,mid),queryMax((i<<1)|1,mid+1,_r));    }}stx[MAXN*4];int n;void build(int i,int l,int r){    stx[i].l = l;    stx[i].r = r;    stx[i].build(1,1,n);    if(l == r)    {        locx[l] = i;        return;    }    int mid = (l+r)/2;    build(i<<1,l,mid);    build((i<<1)|1,mid+1,r);}//修改值void Modify(int x,int y,int val){    int tx = locx[x];    int ty = locy[y];    stx[tx].sty[ty].Min = stx[tx].sty[ty].Max = val;    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].Min = min(stx[i<<1].sty[j].Min,stx[(i<<1)|1].sty[j].Min);                stx[i].sty[j].Max = max(stx[i<<1].sty[j].Max,stx[(i<<1)|1].sty[j].Max);            }            else            {                stx[i].sty[j].Min = min(stx[i].sty[j<<1].Min,stx[i].sty[(j<<1)|1].Min);                stx[i].sty[j].Max = max(stx[i].sty[j<<1].Max,stx[i].sty[(j<<1)|1].Max);            }        }}int queryMin(int i,int x1,int x2,int y1,int y2){    if(stx[i].l == x1 && stx[i].r == x2)        return stx[i].queryMin(1,y1,y2);    int mid = (stx[i].l + stx[i].r)/2;    if(x2 <= mid)return queryMin(i<<1,x1,x2,y1,y2);    else if(x1 > mid)return queryMin((i<<1)|1,x1,x2,y1,y2);    else return min(queryMin(i<<1,x1,mid,y1,y2),queryMin((i<<1)|1,mid+1,x2,y1,y2));}int queryMax(int i,int x1,int x2,int y1,int y2){    if(stx[i].l == x1 && stx[i].r == x2)        return stx[i].queryMax(1,y1,y2);    int mid = (stx[i].l + stx[i].r)/2;    if(x2 <= mid)return queryMax(i<<1,x1,x2,y1,y2);    else if(x1 > mid)return queryMax((i<<1)|1,x1,x2,y1,y2);    else return max(queryMax(i<<1,x1,mid,y1,y2),queryMax((i<<1)|1,mid+1,x2,y1,y2));}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int T;    scanf("%d",&T);    int iCase = 0;    while(T--)    {        iCase++;        printf("Case #%d:\n",iCase);        scanf("%d",&n);        build(1,1,n);        for(int i = 1;i <= n;i++)            for(int j = 1;j <= n;j++)            {                int a;                scanf("%d",&a);                Modify(i,j,a);            }        int q;        int x,y,L;        scanf("%d",&q);        while(q--)        {            scanf("%d%d%d",&x,&y,&L);            int x1 = max(x - L/2,1);            int x2 = min(x + L/2,n);            int y1 = max(y - L/2,1);            int y2 = min(y + L/2,n);            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);            Modify(x,y,t);        }    }    return 0;}