HDU - 1081 To The Max （ 最大子矩阵）

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：最大子矩阵和

解题思路：因为是连续的，所以一切都很简单，稍微思考就可以转化为最大连续子段和的问题了。我们枚举由行组成的连续矩阵，然后对每个矩阵的每一列求和，然后求最大连续子段和就可以了。

#include
#include
#include
#include

using namespace std;
const int MAXN=105;
int dp[MAXN];//保存从0~i的最大和
int a[MAXN][MAXN];
int N;

int temp[MAXN];//暂时保存每一列的和

//求最大连续子段和
int getZDLXZDH(){
    dp[0]=temp[0];
    int maxsum=dp[0];
    for(int i=1;i=0)
            dp[i]=dp[i-1]+temp[i];
        else
            dp[i]=temp[i];

        if(dp[i]>maxsum)
            maxsum=dp[i];

    }
    return maxsum;
}


int main(){


    while(~scanf("%d",&N)){


        for(int i=0;i