hdu 1081 To The Max（dp+化二维为一维）

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): 8839    Accepted Submission(s): 4281

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

 

题目大意：在一个数的矩阵里面找一个和最大矩阵。

题目思路:化二维为一维，转化成1003即可。先求一列一列的和，这样就转化成了一维了。举个例子，我先计算第一行得到dp[0]=0dp[1]=-2,dp[2]=-7,dp[3]=0;这些就是所谓的和，然后在用1003的方法做。那么第二行，就会更新，dp[0]=9,dp[1]=0,dp[2]=-13,dp[3]=2;这些依旧是和，这也就是转换成了一维数组，用1003的方法解决，以此类推~~~

 

详见代码。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 

 5 using namespace std;

 6 

 7 int dp[110];

 8 int num[110][110];

 9 

10 int main ()

11 {

12     int n;

13     while (~scanf("%d",&n))

14     {

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

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

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

18         int ans=-99999;

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

20         {

21             memset(dp,0,sizeof(dp));

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

23             {

24                 int Max=-1;

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

26                 {

27                     dp[k]+=num[j][k];       //先计算出所有的dp和

28                 }

29                 for (int k=0; k<n; k++)     //1003的做法，代码类似

30                 {

31                     if (Max+dp[k]<dp[k])

32                         Max=dp[k];

33                     else

34                         Max=Max+dp[k];

35                     if (ans<Max)              //不断更新最大值

36                     {

37                         ans=Max;

38                         //cout<<j<<" "<<k<<endl;

39                     }

40                 }

41             }

42         }

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

44     }

45     return 0;

46 }