题目是：

SJTUOJ 1002.二哥种花生

二哥在自己的后花园里种了一些花生，也快到了收获的时候了。这片花生地是一个长度为L、宽度为W的矩形，每个单位面积上花生产量都是独立的。他想知道，对于某个指定的区域大小，在这么大的矩形区域内，花生的产量最大会是多少。

首先尝试用最简单的遍历:

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#include <iostream>
 
using namespace std;
 
int cc(int **matp, int m, int l, int h);
 
int main()
{
    int m, n, l, h;
    cin >> m >> n;
    int **matp = new int*[m];
    for (int i = 0; i < m; i++)
    {
        *(matp + i) = new int[n];
    }
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            cin >> matp[i][j];
        }
    }
    cin >> l >> h;
 
    int max = 0;
    int temp = 0;
    for (int i = 0; i < m-l+1; i++)
    {
        for (int j = 0; j < n-h+1; j++)
        {
            temp = cc(matp + i, j, l, h);
            max = (temp >= max) ? temp : max;
        }
    }
    cout << max;
    return 0;
}
 
int cc(int **matp, int xa, int l, int h)
{
    int ret = 0;
    for (int i = 0; i < l; i++)
    {
        for (int j = 0; j < h; j++)
        {
            ret += matp[i][j + xa];
        }
    }
    return ret;
}

发现反馈是6/10 [Time Limit Exceeded]

本身在搜索数组很大的时候，同一个数上要重新计算次，浪费了很多时间。

后来想到，可以采用动窗法来解决。

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#include <iostream>
using namespace std;
 
int main()
{
    
    int mat_h, mat_l;
    cin >> mat_h >> mat_l;
 
    
    
    int **Pmat = new int *[mat_h];
    for (int i = 0; i < mat_h; i++)
        *(Pmat + i) = new int[mat_l];
    int *Pmat_cfg = new int(mat_h);
    
 
    for (int i = 0; i < mat_h; i++)
        for (int j = 0; j < mat_l; j++)
            cin >> Pmat[i][j];
    int srch_l, srch_h;
    cin >> srch_h >> srch_l;
 
    
    
    int **Resmat = new int *[mat_h - srch_h + 1];
    for (int i = 0; i < mat_h - srch_h + 1; i++)
        *(Resmat + i) = new int[mat_l - srch_l + 1]();
    int *Resmat_cfg = new int(mat_h - srch_h + 1);
    
 
    
    for (int i = 0; i < srch_h; i++)
        for (int j = 0; j < srch_l; j++)
            Resmat[0][0] += Pmat[i][j];
 
    
    for (int j = 1; j < mat_l - srch_l + 1; j++)
    {
        Resmat[0][j] = Resmat[0][j - 1];
        for (int i = 0; i < srch_h; i++)
        {
            
            Resmat[0][j] -= Pmat[i][j - 1];
            Resmat[0][j] += Pmat[i][srch_l + j - 1];
        }
    }
 
    
    for (int i = 1; i < mat_h - srch_h + 1; i++)
    {
        
        Resmat[i][0] = Resmat[i - 1][0];
        for (int j = 0; j < srch_l; j++)
        {
            
            Resmat[i][0] -= Pmat[i - 1][j];
            Resmat[i][0] += Pmat[i + srch_h - 1][j];
        }
        
        for (int j = 1; j < mat_l - srch_l + 1; j++)
        {
            Resmat[i][j] = Resmat[i][j - 1];
            for (int z = 0; z < srch_h; z++)
            {
                
                Resmat[i][j] -= Pmat[i + z][j - 1];
                Resmat[i][j] += Pmat[i + z][srch_l + j - 1];
            }
        }
    }
 
    
    int max = 0;
    for (int i = 0; i < mat_h - srch_h + 1; i++)
        for (int j = 0; j < mat_l - srch_l + 1; j++)
            max = max < Resmat[i][j] ? Resmat[i][j] : max;
 
    cout << max;
 
    
    
    
    
    for (int i = 0; i < *Resmat_cfg; i++)
        delete [] *(Resmat + i);
    delete [] Resmat;delete Resmat_cfg;
    
    
    
    
    for (int i = 0; i < *Pmat_cfg; i++)
        delete [] *(Pmat + i);
    delete [] Pmat;delete Pmat_cfg;
    
    
    return 0;
}

仍然是9/10超时。

后来看到，如果不使用原数组，而是建立一个前缀和数组(Cf. FineArtz, 2018)，用来存储从（0,0）到这个点的所有数字和。这样计算取得了比较好的效果。

容易分析得到，动窗法适合一个固定大小的窗户的情形，但是在窗户的大小会发生变化时，需要重新对整个窗户进行计算。在此题中，窗户按不同大小有n^2个，这样做效率会比较低。

而采用前缀和的方式，在经过O(n)的one-pass计算后，每次取值都是O(1)的。

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#include <iostream>

using namespace std;

int main()
{
    int m, n;
    cin >> m >> n;
    
    
    int **Summat = new int *[m + 1];
    for (int i = 0; i < m + 1; i++)
        *(Summat + i) = new int[n + 1]();
    int *Summat_cfg = new int(m + 1);
    
    int getnum;
    for (int i = 1; i <= m; i++)
        for (int j = 1; j <= n; j++)
        {
            cin >> getnum;
            Summat[i][j] = getnum + Summat[i - 1][j] + Summat[i][j - 1] - Summat[i - 1][j - 1];
        }
    int l, h;
    cin >> l >> h;
    int max = 0;
    int total = 0;
    for (int i = 0; i < m + 1 - l; i++)
        for (int j = 0; j < n + 1 - h; j++)
        {
            total = (Summat[i + l][j + h] + Summat[i][j] - Summat[i + l][j] - Summat[i][j + h]);
            max = total > max ? total : max;
        }
    cout << max;
    
    
    for (int i = 0; i < *Summat_cfg; i++)
        delete[] * (Summat + i);
    delete[] Summat;
    delete Summat_cfg;
    
    return 0;
}

这么久重新回来重写这篇文章，其实优化一个算法的核心就是要减少重复，拿空间换时间。（这个题目甚至没有使用更多的空间）有名的Strassen算法，仅仅是减少了一次计算，都能够带来很大的 理论上的收益。(虽然现实意义很小，还不如硬算)

Author: Victrid

Permanent Link: https://victrid.dev/2020/prefix-sum-simple/

License: Copyright (c) 2025 victrid Terms of Use

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