【BZOJ 1047】[HAOI2007]理想的正方形

【链接】 我是链接,点我呀:)

【题意】

在这里输入题意

【题解】

二维的ST表。
每个大的正方形可以由4个小的正方形组成。
然后区域内的最大值最小值。也可以由4个小的张方形部分全部覆盖到。

【代码】

#include <bits/stdc++.h>
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define all(x) x.begin(),x.end()
#define pb push_back
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1
using namespace std;

const double pi = acos(-1);
const int dx[4] = {0,0,1,-1};
const int dy[4] = {1,-1,0,0};
const int N = 1000;
const int M = 10;

int a,b,n;
int ma[N+10][N+10][M+2],mi[N+10][N+10][M+2];

int get_max(int a,int b,int c,int d){
    return max(a,max(b,max(c,d)));
}

int get_min(int a,int b,int c,int d){
    return min(a,min(b,min(c,d)));
}

int main(){
	#ifdef LOCAL_DEFINE
	    freopen("rush_in.txt", "r", stdin);
	#endif
    scanf("%d%d%d",&a,&b,&n);
    rep1(i,1,a)
        rep1(j,1,b){
            scanf("%d",&ma[i][j][0]);
            mi[i][j][0] = ma[i][j][0];
        }
    rep1(k,1,M){
        int len = 1<<k;
        rep1(i,1,a-len+1)
            rep1(j,1,b-len+1){
                ma[i][j][k] = get_max(ma[i][j][k-1],ma[i][j+len/2][k-1],ma[i+len/2][j][k-1],ma[i+len/2][j+len/2][k-1]);
                mi[i][j][k] = get_min(mi[i][j][k-1],mi[i][j+len/2][k-1],mi[i+len/2][j][k-1],mi[i+len/2][j+len/2][k-1]);
            }
    }
    int need = log2(n);
    int ans = -1;
    rep1(i,1,a-n+1)
        rep1(j,1,b-n+1){
            int temp1 = ma[i][j][need],temp2 = ma[i][j+n-(1<<need)][need];
            int temp3 = ma[i+n-(1<<need)][j][need],temp4 = ma[i+n-(1<<need)][j+n-(1<<need)][need];
            temp1 = get_max(temp1,temp2,temp3,temp4);

            int tmp1 = mi[i][j][need],tmp2 = mi[i][j+n-(1<<need)][need];
            int tmp3 = mi[i+n-(1<<need)][j][need],tmp4 = mi[i+n-(1<<need)][j+n-(1<<need)][need];
            tmp1 = get_min(tmp1,tmp2,tmp3,tmp4);
            if (ans==-1){
                ans = temp1-tmp1;
            }else{
                ans = min(ans,temp1-tmp1);
            }
        }
    cout<<ans<<endl;
	return 0;
}