poj 1050 To the Max（动态规划处理二维最大子段和）

1、http://poj.org/problem?id=1050

2、题目大意：

给一个N，然后给定一个N*N的二维数组，然后求一个子矩阵，使得其中的数加起来和最大

3、思路：

将二维数组转换成一维数组，假设二维数组是M行N列，那么将二维数组分成N条，用dp[i]记录第i列的和（可以是任意连续长度，for循环就能实现），那么将dp[i]看做一个一个的数，就转换成了一维的数组

核心代码：

for(int i=1;i<=n;i++)//循环从第一行开始，往下找，i作为每一条的起始点

{

memset(dp,0,sizeof(dp));

for(int j=i;j<=n;j++)

{

for(int k=1;k<=n;k++)

dp[k]+=a[j][k];//求出每一条起点是i，终点是j的长度区间的和

int sum=maxsum(dp,n);//转换成一维数组后，处理就跟一位数组的代码完全一样

if(sum>max)

max=sum;

}

memset(dp,0,sizeof(dp));

}

3、题目：

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 34519		Accepted: 18104

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

4、一遍ac的代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[110][110];
int dp[110];
int b[110];
int maxsum(int *bb,int n)
{
int max=-9999999;
b[0]=max;
for(int i=1;i<=n;i++)
{
if(b[i-1]>0)
b[i]=b[i-1]+bb[i];
if(b[i-1]<0)
b[i]=bb[i];
if(b[i]>max)
max=b[i];
}
return max;
}
int main()
{
int n,max=-99999999;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&a[i][j]);
}
}
for(int i=1;i<=n;i++)
{
memset(dp,0,sizeof(dp));
for(int j=i;j<=n;j++)
{
for(int k=1;k<=n;k++)
dp[k]+=a[j][k];
int sum=maxsum(dp,n);
if(sum>max)
max=sum;
}
memset(dp,0,sizeof(dp));
}
printf("%d\n",max);
return 0;
}
/*
4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2
*/