POJ 1050 To the Max

http://poj.org/problem?id=1050

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. 
As an example, the maximal sub-rectangle of the array: 

0 -2 -7 0 
9 2 -6 2 
-4 1 -4 1 
-1 8 0 -2 
is in the lower left corner: 

9 2 
-4 1 
-1 8 
and has a sum of 15. 

Input

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

150

代码：

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;

const int maxn = 110;
int N;
int mp[maxn][maxn], sum[maxn][maxn];
int maxx = -1e5;

int Maxnum(int l, int r) {
    int c[maxn], dp[maxn];
    for(int i = 0; i < N; i ++)
        c[i] = sum[i][r] - sum[i][l - 1];

    dp[0] = c[0];
    int ans = max(ans, dp[0]);
    for(int i = 1; i < N; i ++) {
        dp[i] = max(dp[i - 1] + c[i], c[i]);
        ans = max(ans, dp[i]);
    }
    return ans;
}

int main() {
    scanf("%d", &N);
    for(int i = 0; i < N; i ++) {
        for(int j = 0; j < N; j ++) {
            scanf("%d", &mp[i][j]);
        }
    }

    for(int j = 0; j < N; j ++) {
        for(int i = 0; i < N; i ++) {
            if(i == 0) sum[j][i] = mp[i][j];
            else sum[j][i] = sum[j][i - 1] + mp[i][j];
        }
    }

    for(int i = 0; i < N; i ++) {
        for(int j = N - 1; j > i; j --) {
            maxx = max(maxx, Maxnum(i, j));
        }
    }

    printf("%d
", maxx);

    return 0;
}

　　枚举上下边界然后枚举最大子序列的值 $O(N^3)$ 时间复杂度