ZOJ 1074 最大和子矩阵

To the Max

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Problem

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.

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.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 

将任意两行之间列元素相加，得到数列。求最大和子矩阵就等价于求最大和子序列。

#include <iostream>
#include <cstring>

using namespace std;

#define MAX 101

int array[MAX][MAX], n, rows_sum[MAX];
int find_best(){
    int sum = 0, max = 0;
    for(int i = 0; i < n; i++){
        sum += rows_sum[i];
        if( sum < 0) sum = 0;
        if( sum > max) max = sum;
    }
    return max;
}
int main(){
    int max;
    while(cin>>n){
        for(int row = 0; row < n; row++)
            for(int col = 0; col < n; col++)
                cin>>array[row][col];
        max = 0;
        for(int start_row = 0; start_row < n; start_row++){
            memset(rows_sum, 0, sizeof(rows_sum));
            for(int row = start_row; row < n; row++){
                for(int col = 0; col < n; col++){
                    rows_sum[col] += array[row][col];
                }
                if( max < find_best()) max = find_best();
            }
        }
        cout<<max;
    }
    return 0;
}