LightOj 1081 二维线段树

LightOj 1081
题目链接：
http://www.bnuoj.com/v3/problem_show.php?pid=13000
题意：
问一个给定矩形中某一块区域的最大值。
思路：
裸二维线段树不带修改。
源码：

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <string>#include <algorithm>#include <iostream>using namespace std;const int MAXN = 501 * 4;int tree[MAXN][MAXN];int a[MAXN/4][MAXN/4];int n, q;void push_upx(int ox, int oy){tree[ox][oy] = max(tree[ox << 1][oy], tree[ox << 1 | 1][oy]);}void push_upy(int ox, int oy){tree[ox][oy] = max(tree[ox][oy << 1], tree[ox][oy << 1 | 1]);}void buildy(int ox, int oy, int ly, int ry, int flag, int lx){//    printf("ox = %d, oy = %d, ly = %d, ry = %d\n", ox, oy, ly, ry);//    system("pause");    if(ly == ry){        if(flag) tree[ox][oy] = a[lx][ly];        else push_upx(ox, oy);    }    else{        int mid = (ly + ry) >> 1;        buildy(ox, oy << 1, ly, mid, flag, lx);        buildy(ox, oy << 1 | 1, mid + 1, ry, flag, lx);        push_upy(ox, oy);    }}void buildx(int ox, int lx, int rx){//    printf("ox = %d, lx = %d, rx = %d\n", ox, lx, rx);//    system("pause");    if(lx != rx){        int mid = (lx + rx) >> 1;        buildx(ox << 1, lx, mid);        buildx(ox << 1 | 1, mid + 1, rx);    }    buildy(ox, 1, 1, n, lx == rx, lx);}int queryy(int ox, int oy, int ly, int ry, int y1, int s){//    printf("ox = %d, oy = %d, ly = %d, ry = %d, y1 = %d, s = %d\n", ox, oy, ly, ry, y1, s);//    system("pause");    int y2 = y1 + s - 1;    if(ly >= y1 && ry <= y2) return tree[ox][oy];    else{        int ans = 0;        int mid = (ly + ry) >> 1;        if(y1 <= mid) ans = max(ans, queryy(ox, oy << 1, ly, mid, y1, s));        if(y2 > mid) ans = max(ans, queryy(ox, oy << 1 | 1, mid + 1, ry, y1, s));        return ans;    }}int queryx(int ox, int lx, int rx, int x1, int y1, int s){//    printf("ox = %d, lx = %d, rx = %d, x1 = %d, y1 = %d, s = %d\n", ox, lx, rx, x1, y1, s);//    system("pause");    int x2 = x1 + s - 1;    if(lx >= x1 && rx <= x2) return queryy(ox, 1, 1, n, y1, s);    else{        int mid = (lx + rx) >> 1;        int ans = 0;        if(x1 <= mid) ans = max(ans, queryx(ox << 1, lx, mid, x1, y1, s));        if(x2 > mid) ans = max(ans, queryx(ox << 1 | 1, mid + 1, rx, x1, y1, s));        return ans;    }}int main(){    int T;    scanf("%d", &T);    for(int cas = 1 ; cas <= T ; cas++){        scanf("%d%d", &n, &q);        for(int i = 1 ; i <= n ; i++) for(int j = 1 ; j <= n ; j++) scanf("%d", &a[i][j]);        buildx(1, 1, n);        printf("Case %d:\n", cas);        for(int i = 1 ; i <= q ; i++){            int u, v, s;            scanf("%d%d%d", &u, &v, &s);            printf("%d\n", queryx(1, 1, n, u, v, s));        }    }    return 0;}