HDU 1081 (DP)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1081

 

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8920    Accepted Submission(s): 4312

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

 

 

题意分析：二维矩阵压缩成一维矩阵，三个for循环实现：

             for (int i=0; i<n; i++)

                   for (int j=i; j<n; j++)  //注意这里的j=i；

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

 

              原理：首先把每一列的最大值求出来，二维就变成的一维，就变成了求最大数列和的问题了，但要注意每次更新。

 

#include <cstdio>
#include <cstring>
using namespace std;

int a[111][111],dp[111],n;
int sum,ans;

int main ()
{
    int i,j,k;
    while (scanf ("%d",&n)==1)
    {
        for (i=0; i<n; i++)
            for (j=0; j<n; j++)
                scanf ("%d",&a[i][j]);
        ans = -99999999;
        for (i=0; i<n; i++)
        {
            memset(dp, 0, sizeof(dp));//每次更新dp清零
            for (j=i; j<n; j++)
            {
                sum = -1;//同上
                for (k=0; k<n; k++)//每列元素相加
                {
                    dp[k] += a[j][k];
                }
                for (k=0; k<n; k++)//找到和最大的那列
                {
                    if (sum > 0)
                        sum += dp[k];
                    else
                        sum = dp[k];
                    if (sum > ans)
                        ans = sum;
                }
            }
        }
        printf ("%d\n",ans);
    }
    return 0;
}