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[][];
 int sum[][];
 int main()
 {
     int nn,i,j,ans,temp;
    // freopen("test.in","r",stdin);
     while(scanf("%d",&nn)!=EOF)
     {
        ans=;
       memset(sum,,sizeof(sum));
       for(i=;i<=nn;i++)
       {
           for(j=;j<=nn;j++)
           {
            scanf("%d",&arr[i][j]);
            sum[i][j]+=arr[i][j]+sum[i][j-];   //求出每一层逐步之和
           }
       }
       for(i=;i<=nn;i++)
       {
           for(j=;j<=nn;j++)
           {
             sum[i][j]+=sum[i-][j];   //在和的基础上，逐步求出最大和
           }
       }
         ans=;
     for(int rr=;rr<=nn;rr++)
     {
       for(int cc=;cc<=nn;cc++)
       {
        for(i=rr;i<=nn;i++)
        {
          for(j=cc;j<=nn;j++)
          {
            temp=sum[i][j]-sum[i][cc-]-sum[rr-][j]+sum[rr-][cc-];
            if(ans<temp)  ans=temp;
          }
        }
       }
     }
     printf("%d\n",ans);
  }
     return ;
 }

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=;
     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,,sizeof(sum));
          for(i=;i<=n;i++)
            for(j=;j<=n;j++)
            {
              scanf("%d",&arr[i][j]);
              sum[j][i]=sum[j][i-]+arr[i][j];
            }
              ans=;
          for(i=;i<=n;i++){

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