UVa 108

　　题目大意：给一个矩阵，找出一个元素之和最大的子矩阵。

　　可以暴力枚举，不过是时间复杂度是O(n⁶)，通过利用一个sum数组保存已计算的和可以提高效率（sum[i][j]保存从矩阵左上角到m[i][j]的所有元素和），把时间复杂度降为O(n⁴)，都说这是DP，可是我怎么感觉不出来？对DP理解不够深刻的原因吗？

#include <cstdio>
#define MAXN 105

int a[MAXN][MAXN], sum[MAXN][MAXN];

int main()
{
#ifdef LOCAL
    freopen("in", "r", stdin);
#endif
    int n;
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                scanf("%d", &a[i][j]);
        for (int i = 0; i <= n; i++) 
            sum[i][0] = sum[0][i] = 0;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                sum[i][j] = sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1] + a[i][j];
        int max = a[1][1];
        for (int i = 1; i <= n; i++)
            for (int p = i; p <= n; p++)
                for (int j = 1; j <= n; j++)
                    for (int q = j; q <= n; q++)
                    {
                        int t = sum[p][q] - sum[p][j-1] - sum[i-1][q] + sum[i-1][j-1];
                        if (t > max)   max = t;
                    }
        printf("%d
", max);
    }
    return 0;
}

 　　还可以进行进一步的优化，将时间复杂度降到O(n³)，主要思想就是将二维转化为一维。先枚举行，比如把子矩阵行范围限制在i行到p行之间，然后求这个行范围内每列的和，这样就转化成了一维的最大连续和问题。

#include <cstdio>
#include <algorithm>
#include <climits>
using namespace std;
#define MAXN 105

int a[MAXN][MAXN], sum[MAXN][MAXN];

int column_sum(int i, int p, int j)
{
    return sum[p][j] - sum[p][j-1] - sum[i-1][j] + sum[i-1][j-1];
}

int main()
{
#ifdef LOCAL
    freopen("in", "r", stdin);
#endif
    int n;
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 0; i <= n; i++) 
            sum[i][0] = sum[0][i] = 0;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
            {
                scanf("%d", &a[i][j]);
                sum[i][j] = sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1] + a[i][j];
            }
        int gMax = a[1][1];
        for (int i = 1; i <= n; i++)
            for (int p = i; p <= n; p++)
            {
                int sum = 0, minSum = 0, lMax = INT_MIN;
                for (int j = 1; j < n; j++)
                {
                    sum += column_sum(i, p, j);
                    lMax = max(lMax, sum - minSum);
                    minSum = min(minSum, sum);
                }
                if (lMax > gMax)  gMax = lMax;
            }    
        printf("%d
", gMax);
    }
    return 0;
}