URAL1146 & POJ1050 Maximum Sum （最大连续子序列和）

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.

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-hand corner and has the sum of 15.

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 white-space (newlines and spaces). These N ²integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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].

The output is the sum of the maximal sub-rectangle.

 

题意解析：

题目比较简单，给定一个N*N的矩阵，求这个矩阵中子矩阵的最大和。譬如给定的示例中，最大的子矩阵为

9	2
−4	1
−1	8

即为15.

 

解题思路：

这个是一个二维的数组，求最大子矩阵和，应该要想到使用一维的最大子序列和，然后就可以将子矩阵压缩为子序列，再求解。

1）将矩阵缩为一列列，然后对这四个列进行 求最大子序列的和。

2）求一个一维的最大子序列和。

 

代码如下：

#include <stdio.h>
#define MAX_N 100
int arr[MAX_N][MAX_N];
int rowsSum[MAX_N]; //每一行的和
int N;
int calc(int x, int y);
int main()
{
      //freopen("input.txt","r",stdin);
      scanf("%d",&N);
      int i = 0;
      int j = 0;
      int max = -1270000;
      int sum = 0;
      for(i=0;i<N;i++)
      {
             for(j=0;j<N;j++)
             {
                   scanf(" %d",&arr[i][j]);
             }
      }

      for(i=0;i<N;i++)
      {
           for(j=i;j<N;j++)
          {
                 sum = calc(i,j); //计算第i列到第j列的最大和
                 if(sum>max)
                {
                      max = sum;
                }
          }
      }
      printf("%d\n",max);
      return 0;
}

int calc(int x, int y)
{
      int i = 0;
      int j = 0;
      int resultValue = -1270000;
      int thisSum = 0;
      for(i=0;i<N;i++)
      {
            rowsSum[i] = 0;
            for(j=x;j<=y;j++)
            {
                   rowsSum[i] = rowsSum[i] + arr[i][j];
            }
      }

      //rowsSum表示，对于指定的列i到j，每一行的和。

      //求rowsSum[]这个一维数组的最大子序列和
      for(i=0;i<N;i++)
      {
            thisSum = thisSum + rowsSum[i];
            if(thisSum>resultValue)
            {
                    resultValue = thisSum;
            }
            if(thisSum<0)
            {
                    thisSum = 0;
            }
      }

       return resultValue;

}