POJ 1050 To the Max

最新推荐文章于 2021-10-04 16:49:11 发布

原创最新推荐文章于 2021-10-04 16:49:11 发布 · 290 阅读

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本文介绍了一种解决二维数组中寻找具有最大和的子矩阵的方法。通过暴力枚举行并利用一维最大子数组求和算法，该算法能够有效地找到最大子矩阵及其和。

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.

【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。

【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);
}