如何在高度图中找到体积最大的长方体？ （低复杂度）

Question

I need to find the cuboid with the greatest volume, contained within a 2D-heightmap. The heightmap is an array of size w*d where w is width, h is height and d is depth. In C, this would look along the lines of:

unsigned heightmap[w][d]; // all values are <= h

I already know that there is a naive algorithm which can solve this with O(w*d*h) complexity. However, I suspect that there is a more optimal method out there. It works as follows, in pythonic pseudocode:

resultRectangle = None
resultHeight = None
resultVolume = -1

# iterate over all heights
for loopHeight in range(0, h):
    # create a 2D bitmap from our heightmap where a 1 represents a height >= loopHeight
    bool bitmap[w][d]
    for x in range(0, w):
        for y in range(0, d):
            bitmap[x][y] = heightmap[x][y] >= loopHeight

    # obtain the greatest-volume cuboid at this particular height
    maxRectangle = maxRectangleInBitmap(bitmap)
    volume = maxRectangle.area() * loopHeight

    # compare it to our current maximum and replace it if we found a greater cuboid
    if volume > resultVolume:
        resultHeight = loopHeight
        resultVolume = volume
        resultRectangle = maxRectangle

resultCuboid = resultRectangle.withHeight(resultHeight)

Finding the greatest area of all 1 in a rectangle is a known problem with O(1) complexity per pixel or O(w*d) in our case. The total complexity of the naive approach is thus O(w*h*d) .

So as I already stated, I was wondering if we can beat this complexity. Perhaps we can get it down to O(w*d * log(h)) by searching through heights more intelligently instead of "brute-forcing" all of them.

The answer to this question Find largest cuboid containing only 1's in an NxNxN binary array by Evgeny Kluev seems to take a similar approach, but it falsely(?) assumes that the volumes which we would find at these heights form a unimodal function. If this was the case, we could use Golden Section Search to choose heights more intelligently, but I don't think we can.

Answer 1

Here is an idea, with a significant assumption. pseudo-code:

P <- points from heightmap sorted by increasing height.
R <- set of rectangles. All maximal empty sub-rectangles for the current height.
R.add(Rectangle(0,0,W,H)
result = last_point_in(P).height()
foreach(p in P):
   RR <- rectangles from R that overlap P (can be found in O(size(RR)), possibly with some logarithmic factors)
   R = R - RR
   foreach(r in RR)
       result = max(result, r.area() * p.height())
       split up r, adding O(1) new rectangles to R.
return result

The assumption, which I have a gut feeling about, but can't prove, is that RR will be O(1) size on average.

Edit: to clarify the "splittting", if we split at point p:

AAAAADFFF
AAAAADFFF
AAAAADFFF
BBBBBpGGG
CCCCCEHHH
CCCCCEHHH

We generate new rectangles consisting of: ABC, CEH, FGH, ADF, and add them to R.

Answer 2

OK, another take. Most "meat" is in the go function. It uses the same "splitting" concept as in my other answer, but uses top-down dynamic programming with memoization. rmq2d implements 2D Range Minimum Query. for size 1000x1000 it takes about 30 seconds (while using 3GB of memory).

#include <iostream>
#include <vector>
#include <cassert>
#include <set>
#include <tuple>
#include <memory.h>
#include <limits.h>

using namespace std;

constexpr int ilog2(int x){
    return 31 - __builtin_clz(x);
}

const int MAX_DIM = 100;




template<class T>
struct rmq2d{
    struct point{
        int x,y;

        point():x(0),y(0){}
        point(int x,int y):x(x),y(y){}
    };
    typedef point array_t[MAX_DIM][ilog2(MAX_DIM)+1][MAX_DIM];

    int h, logh;
    int w, logw;
    vector<vector<T>> v;

    array_t *A;

    rmq2d(){A=nullptr;}

    rmq2d &operator=(const rmq2d &other){
        assert(sizeof(point)==8);
        if(this == &other) return *this;
        if(!A){
            A = new array_t[ilog2(MAX_DIM)+1];
        }
        v=other.v;
        h=other.h;
        logh = other.logh;
        w=other.w;
        logw=other.logw;
        memcpy(A, other.A, (ilog2(MAX_DIM)+1)*sizeof(array_t));
        return *this;
    }

    rmq2d(const rmq2d &other){
        A = nullptr;
        *this = other;
    }

    ~rmq2d(){
        delete[] A;
    }


    T query(point pos){
        return v[pos.y][pos.x];
    }

    rmq2d(vector<vector<T>> &v) : v(v){
        A = new array_t[ilog2(MAX_DIM)+1];
        h = (int)v.size();
        logh = ilog2(h) + 1;
        w = (int)v[0].size();
        logw = ilog2(w) + 1;

        for(int y=0; y<h; ++y){
            for(int x=0;x<w;x++) A[0][y][0][x] = {x, y};

            for(int jx=1; jx<logw; jx++){
                int sz = 1<<(jx-1);
                for(int x=0; x+sz < w; x++){
                    point i1 = A[0][y][jx-1][x];
                    point i2 = A[0][y][jx-1][x+sz];
                    if(query(i1) < query(i2)){
                        A[0][y][jx][x] = i1;
                    }else{
                        A[0][y][jx][x] = i2;
                    }
                }
            }
        }
        for(int jy=1; jy<logh; ++jy){
            int sz = 1<<(jy-1);
            for(int y=0; y+sz<h; ++y){
                for(int jx=0; jx<logw; ++jx){
                    for(int x=0; x<w; ++x){
                        point i1 = A[jy-1][y][jx][x];
                        point i2 = A[jy-1][y+sz][jx][x];
                        if(query(i1) < query(i2)){
                            A[jy][y][jx][x] = i1;
                        }else{
                            A[jy][y][jx][x] = i2;
                        }

                    }
                }
            }
        }
    }

    point pos_q(int x1, int x2, int y1, int y2){
        assert(A);
        int lenx = ilog2(x2 - x1);
        int leny = ilog2(y2 - y1);

        point idxs[] = {
                 A[leny][y1][lenx][x1],
                 A[leny][y2-(1<<leny)][lenx][x1],
                 A[leny][y1][lenx][x2-(1<<lenx)],
                 A[leny][y2-(1<<leny)][lenx][x2-(1<<lenx)]
                };
        point ret = idxs[0];
        for(int i=1; i<4; ++i){
            if(query(ret) > query(idxs[i])) ret = idxs[i];
        }
        return ret;

    }

    T val_q(int x1, int x2, int y1, int y2){
        point pos = pos_q(x1,x2,y1,y2);
        return v[pos.y][pos.x];
    }
};

rmq2d<long long> rmq;

set<tuple<int, int, int ,int>> cac;
vector<vector<long long>> v(MAX_DIM-5,vector<long long>(MAX_DIM-5,0));

long long ret = 0;
int nq = 0;

void go(int x1, int x2, int y1, int y2){
    if(x1 >= x2 || y1>=y2) return;

    if(!cac.insert(make_tuple(x1,y1,x2,y2)).second) return;
    ++nq;

    auto p = rmq.pos_q(x1, x2, y1, y2);
    long long cur = v[p.y][p.x]*(x2-x1)*(y2-y1);
    if(cur > ret){
        cout << x1 << "-" << x2 << ", " << y1 << "-" << y2 << " h=" << v[p.y][p.x] <<  " :" << cur << endl;
        ret = cur;
    }

    go(p.x+1, x2, y1, y2);
    go(x1, p.x, y1, y2);
    go(x1, x2, p.y+1, y2);
    go(x1, x2, y1, p.y);
}


int main(){
    int W = (int)v[0].size();
    int H=(int)v.size();
    for(int y=0; y<H;++y){
        for(int x=0; x<W; ++x){
            v[y][x] = rand()%10000;
        }
    }
    rmq = rmq2d<long long>(v);
    go(0,W, 0, H);
    cout << "nq:" << nq << endl;
}