HDU 4819 二维线段树

HDU 4819
题目链接：
http://www.bnuoj.com/v3/problem_show.php?pid=35690
题意：
给一个初始矩阵，矩阵大小1000 * 1000。
现在有一个操作询问以(i,j)为中心的正方形（保证边长为奇数）中矩阵格子里的最大值和最小值，输出(max+min)/2，并且把这个格子变成这个值。
思路：
裸二维线段树，其实是自己想的写法居然和大多数版差不多……这个版写的还是太繁琐，建议直接用大白书上的版。
源码：

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <iostream>#include <string>#include <vector>using namespace std;#define inf (1000000007)typedef pair<int,int> pii;const int MAXN = 800 + 5;int mmin[MAXN * 4][MAXN * 4], mmax[MAXN * 4][MAXN * 4];int a[MAXN][MAXN];int n;int temp;pii change(pii a, pii b){    pii ans;    ans.first = min(a.first, b.first);    ans.second = max(a.second, b.second);    return ans;}void push_upx(int ox, int oy){    mmin[ox][oy] = min(mmin[ox << 1][oy], mmin[(ox << 1) + 1][oy]);    mmax[ox][oy] = max(mmax[ox << 1][oy], mmax[(ox << 1) + 1][oy]);}void push_upy(int ox, int oy){    mmin[ox][oy] = min(mmin[ox][oy << 1], mmin[ox][(oy << 1) + 1]);    mmax[ox][oy] = max(mmax[ox][oy << 1], mmax[ox][(oy << 1) + 1]);}void buildy(int ox, int lx, int rx, int oy, int ly, int ry){    temp = max(temp, oy);    if(lx == rx){        if(ly == ry) mmin[ox][oy] = mmax[ox][oy] = a[lx][ly];        else{            int mid = (ly + ry) >> 1;            buildy(ox, lx, rx, (oy << 1), ly, mid);            buildy(ox, lx, rx, (oy << 1) + 1, mid + 1, ry);            push_upy(ox, oy);        }    }    else{        if(ly == ry) push_upx(ox, oy);        else{            int mid = (ly + ry) >> 1;            buildy(ox, lx, rx, (oy << 1), ly, mid);            buildy(ox, lx, rx, (oy << 1) + 1, mid + 1, ry);            push_upx(ox, oy);        }    }}void buildx(int ox, int lx, int rx){    if(lx != rx){        int mid = (lx + rx) >> 1;        buildx(ox << 1, lx, mid);        buildx((ox << 1) + 1, mid + 1, rx);    }    buildy(ox, lx, rx, 1, 1, n);}void updatey(int ox, int lx, int rx, int oy, int ly, int ry, int y, int val){//    printf("updatey ox = %d, oy = %d, ly = %d, ry = %d, y = %d, val = %d\n", ox, oy, ly, ry, y, val);    if(lx == rx){        if(ly == ry) mmin[ox][oy] = mmax[ox][oy] = val;        else{            int mid = (ly + ry) >> 1;            if(y <= mid) updatey(ox, lx, rx, oy << 1, ly, mid, y, val);            else updatey(ox, lx, rx, (oy << 1) + 1, mid + 1, ry, y, val);            push_upy(ox, oy);        }    }    else{        if(ly == ry) push_upx(ox, oy);        else{            int mid = (ly + ry) >> 1;            if(y <= mid) updatey(ox, lx, rx, oy << 1, ly, mid, y, val);            else updatey(ox, lx, rx, (oy << 1) + 1, mid + 1, ry, y, val);            push_upy(ox, oy);        }    }}void updatex(int ox, int lx, int rx, int x, int y, int val){//    printf("udpatex ox = %d, lx = %d, rx = %d, x = %d, y = %d, val = %d\n", ox, lx, rx, x, y, val);//    if(val == 6) printf("ox = %d, lx = %d, rx = %d")    if(lx == rx){        updatey(ox, lx, rx, 1, 1, n, y, val);    }    else{        int mid = (lx + rx) >> 1;        if(x <= mid) updatex(ox << 1, lx, mid, x, y, val);        else updatex((ox << 1) + 1, mid + 1, rx, x, y, val);        updatey(ox, lx, rx, 1, 1, n, y, val);    }}pii queryy(int ox, int oy, int ly, int ry, int y, int L){    pii ans = make_pair(inf, -inf);    int y1 = y - L / 2; int y2 = y + L / 2;    if(ly >= y1 && ry <= y2) ans = make_pair(mmin[ox][oy], mmax[ox][oy]);    else{        int mid = (ly + ry) >> 1;        if(y1 <= mid) ans = change(ans, queryy(ox, oy << 1, ly, mid, y, L));        if(y2 > mid) ans = change(ans, queryy(ox, (oy << 1) + 1, mid + 1, ry, y, L));    }    return ans;}pii queryx(int ox, int lx, int rx, int x, int y, int L){    pii ans = make_pair(inf, -inf);    int x1 = x - L / 2; int x2 = x + L / 2;    if(lx >= x1 && rx <= x2) ans = queryy(ox, 1, 1, n, y, L);    else{        int mid = (lx + rx) >> 1;        if(x1 <= mid) ans = change(ans, queryx(ox << 1, lx, mid, x, y, L));        if(x2 > mid) ans = change(ans, queryx((ox << 1) + 1, mid + 1, rx, x, y, L));    }    return ans;}int main(){    int T;    scanf("%d", &T);    for(int cas = 1 ; cas <= T ; cas++){        scanf("%d", &n);        for(int i = 1 ; i <= n ; i++) for(int j = 1 ; j <= n ; j++) scanf("%d", &a[i][j]);        temp = 0;        buildx(1, 1, n);        int q;        printf("Case #%d:\n", cas);        scanf("%d", &q);        while(q--){            int u, v, L;            scanf("%d%d%d", &u, &v, &L);            pii ans = queryx(1, 1, n, u, v, L);//            printf("ans.first = %d, ans.second = %d\n", ans.first, ans.second);            a[u][v] = (ans.second + ans.first) / 2;            printf("%d\n", a[u][v]);            updatex(1, 1, n, u, v, a[u][v]);//            for(int i = 1 ; i <= 5 ; i++){//                for(int j = 1 ; j <= 5 ; j++) printf("%d ", mmax[i][j]);//                printf("\n");//            }//            printf("\n");//            for(int i = 1 ; i <= 5 ; i++){//                for(int j = 1 ; j <= 5 ; j++) printf("%d ", mmin[i][j]);//                printf("\n");//            }        }    }    return 0;}