hdu-------1081To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7681    Accepted Submission(s): 3724

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

 

Source

Greater New York 2001

这道题个人觉得是到灰常好的题目，适合不同层次的人做，求的是最大连续子矩阵和....

方法一：

简单的模拟即可。首先求出每个阶段只和，然后得到这个状态，然后进行矩阵规划，求最大值即可

代码浅显易懂：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
using namespace std;
int arr[101][101];
int sum[101][101];
int main()
{
    int nn,i,j,ans,temp;
   // freopen("test.in","r",stdin);
    while(scanf("%d",&nn)!=EOF)
    {
       ans=0;
      memset(sum,0,sizeof(sum));
      for(i=1;i<=nn;i++)
      {
          for(j=1;j<=nn;j++)
          {
           scanf("%d",&arr[i][j]);
           sum[i][j]+=arr[i][j]+sum[i][j-1];   //求出每一层逐步之和
          }
      }  
      for(i=2;i<=nn;i++)
      {
          for(j=1;j<=nn;j++)
          {
            sum[i][j]+=sum[i-1][j];   //在和的基础上，逐步求出最大和
          }
      }
        ans=0;
    for(int rr=1;rr<=nn;rr++)
    {
      for(int cc=1;cc<=nn;cc++)
      {
       for(i=rr;i<=nn;i++)
       {
         for(j=cc;j<=nn;j++)
         {
           temp=sum[i][j]-sum[i][cc-1]-sum[rr-1][j]+sum[rr-1][cc-1];  
           if(ans<temp)  ans=temp;
         }
       }
      }
    }
    printf("%d
",ans);
 }
    return 0;
}

 方法二：

采用状态压缩的方式进行DP求解最大值......!

代码：

    /*基于一串连续数字进行求解的扩展*/
    /*coder Gxjun 1081*/
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    using namespace std;
    const int nn=101;
    int arr[nn][nn],sum[nn][nn];
    int main()
    {
       int n,i,j,maxc,res,ans;
       //  freopen("test.in","r",stdin);
       while(scanf("%d",&n)!=EOF)
       {
          memset(sum,0,sizeof(sum));
         for(i=1;i<=n;i++)
           for(j=1;j<=n;j++)
           {
             scanf("%d",&arr[i][j]);
             sum[j][i]=sum[j][i-1]+arr[i][j];
           }
             ans=0;
         for(i=1;i<=n;i++){

           for(j=i;j<=n;j++)
           {
               res=maxc=0;
              for(int k=1;k<=n;k++)
              {
                 maxc+=sum[k][j]-sum[k][i-1];
                if(maxc<=0)  maxc=0;
                else
                 if(maxc>res) res=maxc;
              }
                if(ans<res)  ans=res;
           }
         }
         printf("%d
",ans);
       }
        return 0;
    }