【题意】给了一个n*n的矩阵,有两种操作,一种是更新矩形里面某个点的值,一种是查询一个子矩阵的最大和最小值。

【解题方法】2D线段树裸题了。直接上代码了。

//
//Created by just_sort 2016/9/16 14:25
//Copyright (c) 2016 just_sort.All Rights Reserved
//

//2D segment tree.
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 2010;
struct segmentTree2D{
    int mx[maxn][maxn],mi[maxn][maxn],n,m;
    int xo,xleaf,x1,y1,x2,y2,x,y,v,vmax,vmin;
    void query1D(int o,int l,int r)
    {
        if(y1<=l && r<=y2)
        {
            vmax = max(vmax,mx[xo][o]);
            vmin = min(vmin,mi[xo][o]);
        }
        else
        {
            int mid = (l+r)/2;
            if(y1<=mid) query1D(o*2,l,mid);
            if(mid<y2)  query1D(o*2+1,mid+1,r);
        }
    }
    void query2D(int o,int l,int r)
    {
        if(x1<=l && r<=x2)
        {
            xo = o;
            query1D(1,1,m);
        }
        else
        {
            int mid = (l+r)/2;
            if(x1<=mid) query2D(o*2,l,mid);
            if(mid<x2)  query2D(o*2+1,mid+1,r);
        }
    }
    void modify1D(int o,int l,int r)
    {
        if(l==r)
        {
            if(xleaf)
            {
                mx[xo][o]=mi[xo][o]=v;
                return ;
            }
            mx[xo][o] = max(mx[xo*2][o],mx[xo*2+1][o]);
            mi[xo][o] = min(mi[xo*2][o],mi[xo*2+1][o]);
        }
        else
        {
            int mid = (l+r)/2;
            if(y<=mid) modify1D(o*2,l,mid);
            else  modify1D(o*2+1,mid+1,r);
            mx[xo][o] = max(mx[xo][o*2],mx[xo][o*2+1]);
            mi[xo][o] = min(mi[xo][o*2],mi[xo][o*2+1]);
        }
    }
    void modify2D(int o,int l,int r)
    {
        if(l==r)
        {
            xo = o;
            xleaf = 1;
            modify1D(1,1,m);
        }
        else
        {
            int mid = (l+r)/2;
            if(x<=mid) modify2D(o*2,l,mid);
            else modify2D(o*2+1,mid+1,r);
            xo = o;
            xleaf = 0;
            modify1D(1,1,n);
        }
    }
    void query()
    {
        vmax = -inf;
        vmin = inf;
        query2D(1,1,n);
    }
    void modify()
    {
        modify2D(1,1,n);
    }
}t;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        t.n=n,t.m=n;
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                scanf("%d",&t.v);
                t.x=i,t.y=j;
                t.modify();
            }
        }
        int q;
        char op[10];
        scanf("%d",&q);
        while(q--)
        {
            scanf("%s",op);
            if(op[0]=='q')
            {
                scanf("%d%d%d%d",&t.x1,&t.y1,&t.x2,&t.y2);
                t.query();
                printf("%d %d\n",t.vmax,t.vmin);
            }
            else{
                scanf("%d%d%d",&t.x,&t.y,&t.v);
                t.modify();
            }
        }
    }
    return 0;
}