poj 1050 To the Max(最大子矩阵之和，基础DP题)

To the Max

Time Limit: 1000MS
Memory Limit: 10000K

Total Submissions: 38573
Accepted: 20350

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

题解：假设已经知道矩形的上下边界，比如知道矩形的区域的上下边界分别是第a行和第c行，现在要确定左右边界；

代码：

#include <iostream>
#define INF 2147483647
using namespace std;
int a[1010][1010];
int sum[1010][1010];//数组我开的比较大，这无所谓
 
int Maxsum(int n,int m)
{//求最大子矩阵之和
    int i,j,k;
    int Max=-INF;
    for(i=0;i<=n;i++)
        sum[i][0]=0;
    for(i=1;i<=m;i++)
        sum[0][i]=0;
    for(i=1;i<=n;i++)
    {
        for(j=i;j<=n;j++)
        {
            sum[j][m]=sum[j-1][m]+a[j][m];//sum[a][b]储存 第b列中 第1行到第a行之间所有元素的和
            int tmp=sum[j][m]-sum[i-1][m];//此时tmp值为 第m列中 第i行到第j行之间所有元素之和
            int big=tmp;
            for(k=m-1;k>=1;k--)
            {
                if(tmp<0)
                    tmp=0;
                sum[j][k]=sum[j-1][k]+a[j][k];
                tmp+=sum[j][k]-sum[i-1][k];
                big=max(big,tmp);
                Max=max(big,Max);
            }
        }
    }
    return Max;
}
 
int main()
{
    int n,m,i,j;
    cin>>n;
    m=n;//这里可变为cin>>m，则矩阵是n*m
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            cin>>a[i][j];
    cout<<Maxsum(n,m)<<endl;
    return 0;
}