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

最新推荐文章于 2019-07-28 01:42:13 发布

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最新推荐文章于 2019-07-28 01:42:13 发布

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原文链接：http://www.cnblogs.com/xuxu-ning/p/4973623.html

本文介绍了一种寻找二维数组中最大子矩阵和的算法。通过将二维问题转化为一系列一维最大子序列和的问题来解决。文章给出了具体实现代码，并解释了输入输出格式。

Description

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.

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 N² integers separated by white-space (newlines and spaces). These N²integers 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].

Output

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;

}

转载于:https://www.cnblogs.com/xuxu-ning/p/4973623.html