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：

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<cstdlib>

 4 #include<iostream>

 5 using namespace std;

 6 int arr[101][101];

 7 int sum[101][101];

 8 int main()

 9 {

10     int nn,i,j,ans,temp;

11    // freopen("test.in","r",stdin);

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

13     {

14        ans=0;

15       memset(sum,0,sizeof(sum));

16       for(i=1;i<=nn;i++)

17       {

18           for(j=1;j<=nn;j++)

19           {

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

21            sum[i][j]+=arr[i][j]+sum[i][j-1];   //求出每一层逐步之和

22           }

23       }  

24       for(i=2;i<=nn;i++)

25       {

26           for(j=1;j<=nn;j++)

27           {

28             sum[i][j]+=sum[i-1][j];   //在和的基础上，逐步求出最大和

29           }

30       }

31         ans=0;

32     for(int rr=1;rr<=nn;rr++)

33     {

34       for(int cc=1;cc<=nn;cc++)

35       {

36        for(i=rr;i<=nn;i++)

37        {

38          for(j=cc;j<=nn;j++)

39          {

40            temp=sum[i][j]-sum[i][cc-1]-sum[rr-1][j]+sum[rr-1][cc-1];  

41            if(ans<temp)  ans=temp;

42          }

43        }

44       }

45     }

46     printf("%d\n",ans);

47  }

48     return 0;

49 }

View Code

 方法二：

采用状态压缩的方式进行DP求解最大值......!

代码：

 

 1     /*基于一串连续数字进行求解的扩展*/

 2     /*coder Gxjun 1081*/

 3     #include<iostream>

 4     #include<cstdio>

 5     #include<cstring>

 6     #include<cstdlib>

 7     using namespace std;

 8     const int nn=101;

 9     int arr[nn][nn],sum[nn][nn];

10     int main()

11     {

12        int n,i,j,maxc,res,ans;

13        //  freopen("test.in","r",stdin);

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

15        {

16           memset(sum,0,sizeof(sum));

17          for(i=1;i<=n;i++)

18            for(j=1;j<=n;j++)

19            {

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

21              sum[j][i]=sum[j][i-1]+arr[i][j];

22            }

23              ans=0;

24          for(i=1;i<=n;i++){

25 

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

27            {

28                res=maxc=0;

29               for(int k=1;k<=n;k++)

30               {

31                  maxc+=sum[k][j]-sum[k][i-1];

32                 if(maxc<=0)  maxc=0;

33                 else

34                  if(maxc>res) res=maxc;

35               }

36                 if(ans<res)  ans=res;

37            }

38          }

39          printf("%d\n",ans);

40        }

41         return 0;

42     }

View Code