F - 最大子矩形
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Submit

Status
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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

<span style="color:#3333ff;">/*_________________________________________________________________________________________________________________       author      :       Grant Yuan       time        :       2014.7.19       source      :       Hdu1081       algorithm   :       暴力枚举+动态规划       explain     :       暴力枚举第k1行到第k2行每一列各个元素的和sum[i]，然后对sum[i]进行一次最大子序列求和___________________________________________________________________________________________________________________*/#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<functional>#include<algorithm>using namespace std;int a[105][105];int l;int sum[105];int dp[105];int main(){	while(~scanf("%d",&l)){		for(int i=0;i<l;i++)			for(int j=0;j<l;j++)			 scanf("%d",&a[i][j]);		int ans=-9999999,ans1;		for(int k1=0;k1<l;k1++)			for(int k2=0;k2<l;k2++){		   memset(dp,0,sizeof(dp));		   memset(sum,0,sizeof(sum));		 for(int i=0;i<l;i++){			for(int j=k1;j<=k2;j++)		     {			   sum[i]+=a[i][j];		         }		   for(int m=0;m<l;m++)		     {		   	  dp[m+1]=max(dp[m]+sum[m],sum[m]);		      }		   ans1=dp[1];		   for(int d=1;d<=l;d++)			  if(dp[d]>ans1)		ans1=dp[d];		if(ans1>ans)		      ans=ans1;}}		printf("%d\n",ans);}}</span>