HDU 1081 To The Max

Problem Description：

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

题意：有一个n*n的矩阵，里面有负数有正数，现在问该矩阵的子矩阵和最大是多少，子矩阵大于等于1*1。

   原矩阵               Map数组

0 -2 -7 0            0 -2 -9 -9

9 2 -6 2             9 11  5  7

-4 1 -4 1           -4 -3 -7 -6

-1 8 0 -2           -1  7  7  5

可以查找到2~4行第1~2列和最大，它们的和是11+（-3）+ 7 = 15，该子矩阵是：

 9 2

-4 1

-1 8

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

const int INF=0x3f3f3f3f;
const int N=110;

int main ()
{
    int n, i, j, k, Max, sum, Map[N][N], num;

    while (scanf("%d", &n) != EOF)
    {
        memset(Map, 0, sizeof(Map));

        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
            {
                scanf("%d", &num);
                Map[i][j] = Map[i][j-1] + num; ///保存第i行前j列的和
            }
        }

        Max = -INF;

        for (i = 1; i <= n; i++) ///列
        {
            for (j = i; j <= n; j++) ///列 
            {
                sum = 0;

                for (k = 1; k <= n; k++) ///行
                {
                    if (sum < 0) sum = 0; ///sum小于0时加上其它的数会变小，我们要找最大的和，所以重新置为0

                    sum += Map[k][j]-Map[k][i-1]; ///sum记录前k行第i~j列的和

                    if (sum > Max) Max = sum; ///Max保存最大值
                }
            }
        }

        printf("%d
", Max);
    }

    return 0;
}