poj 动态规划之1050 To the Max

poj 1050 To the Max

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

15

解释：为了避免计算量，设置sum[i][j]代表第i行1到j列的和，其余的就是正常的dp

#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>
#include<memory.h>
using namespace std;
int sum[105][105], a[105][105], dp[105];
int main()
{
int N, i, j, k, ans;
ans = 0;
scanf("%d", &N);
memset(sum, 0, sizeof(sum));
memset(dp, 0, sizeof(dp));
for (i = 1; i <= N; i++)
for (j = 1; j <= N; j++)
{
scanf("%d", &a[i][j]);
sum[i][j] = sum[i][j - 1] + a[i][j];
}
for (i = 1; i <= N;i++)
for (j = i; j <= N; j++)
{
dp[1] = sum[1][j] - sum[1][i - 1];
for (k = 2; k <= N; k++)
{
dp[k] = sum[k][j] - sum[k][i - 1];
if (dp[k - 1] > 0)
dp[k] += dp[k - 1];
ans = max(ans, dp[k]);
}
}
printf("%d", ans);
return 0;
}