DP动态规划——（hdu 1081）To the Max

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): 4873    Accepted Submission(s): 2302

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

40 -2 -7 0 
9 2 -6 2-4 1 -4 1 
-1 8 0 -2

 

Sample Output

15

又是一道经典的DP问题，初次见到我想大概都会想到暴力，毕竟我也是这么想的，但却是暴力肯定是不能解决问题的。

这道题是求二维子数组之和的最大值，就是在这个正方形里面取一个矩形，使得得到的数最大。

对于二维的求最大值，同样要知道求一维的最大值。（参考：DP动态规划——最大数字子串：http://blog.csdn.net/someday7_toi/article/details/7852448   ）。

算法的复杂度为O( n^3 )  .

三层循环：

最外层  i 循环 n-1，表明子矩阵是从第 i 列 开始累加的。(实际上为 i+1 )

第二层 j 循环 n，表明子矩阵是从第 i 列累加到第 j 列。

第三层 k 从1 到 N 对行数 做一维 DP。（这样就保证了求出来的是矩形，即子矩阵）
 

b [ i ][ j ]存的是第j列中前i行的数据之和.求 m 到 n 列的和 即b[k][n]-b[k][m-1].

源代码：

#include<stdio.h>#include<string.h>#define M 105int b[M][M];//  b [ i ][ j ]存的是第j列中前i行的数据之和.求 m 到 n 列的和 即b[k][n]-b[k][m-1]int main(){int i,j,k,sum,ans,s;int n;while(~scanf("%d",&n)){ memset(b,0,sizeof(b));ans=0;for(i=1;i<=n;i++)for(j=1;j<=n;j++){scanf("%d",&k);b[i][j]=b[i][j-1]+k;} for(i=0;i<n;i++) //代码为了方便 i 从 0 开始{for(j=i+1;j<=n;j++){sum=0;for(k=1;k<=n;k++) //利用一维DP的方法求最大字串{s=b[k][j]-b[k][i];if(sum+s<0){sum=0;}elsesum+=s;if(sum>ans)ans=sum;}}}printf("%d\n",ans);}return 0;}

 

可以参考：

http://www.cnblogs.com/blackcruiser/articles/1786888.html
http://blog.csdn.net/kindlucy/article/details/5675202