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;
}