区间Dp 暴力枚举+动态规划 Hdu1081

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>