POJ 1050 To the Max

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.

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【题目分析】
与其说是动态规划，其实暴力枚举的成分更多一些，每次枚举一定的行，然后从前往后扫一遍。其实就是最大子序列的和。dp[i]=max(dp[i],dp[i-1]+a[i]);然后暴力的枚举一下。
就是一道水题。

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【代码】

#include <cstdio>#include <cstring>#include <iostream> using namespace std;int a[101][101],n,ans=0;int main(){    cin>>n;    for (int i=1;i<=n;++i)        for (int j=1;j<=n;++j)            cin>>a[i][j],a[i][j]+=a[i-1][j];    for (int i=1;i<=n;++i)        for (int j=i;j<=n;++j)        {            int sum=0;            for (int k=1;k<=n;++k)            {                sum+=a[j][k]-a[i-1][k];                if (sum<0) sum=0;                ans=max(ans,sum);            }        }    printf("%d\n",ans);}