To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7620 Accepted Submission(s): 3692

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

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1
8 0 -2

Sample Output

　　这道题目是求二维数组的最大子矩阵的和，最大子矩阵一定是在1~n行之间，所以要任选连续的几行压缩成一位数组求最大连续子段和。

代码:

#include <iostream>

#include <cstdio>

using namespace std;

#define N 105

int arr[N][N],b[N];

int dp(int *a,int m)    //求一维数组的最大子段和

{

    int i,sum,max;

    sum = 0;

    max = 0;

    for(i=0; i<N; i++)

    {

        sum += a[i];

        if(sum<0)

            sum = 0;

        if(sum>max)

            max = sum;

    }

    return max;

}

int main()

{

    int i,j,k,n,sum,max;

    while(scanf("%d",&n)!=EOF)

    {

        for(i=0; i<n; i++)

            for(j=0; j<n; j++)

                scanf("%d",&arr[i][j]);

        max = 0;

        for(i=0; i<n; i++)

        {

            memset(b,0,sizeof(b));

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

            {

                for(k=0; k<n; k++)

                    b[k] += arr[j][k];

                sum = dp(b,n);

                if(sum>max)

                    max = sum;

            }

        }

        printf("%d\n",max);

    }

    return 0;

}

巴特西

Hdu 1081 To The Max