POJ 1050 To the Max【DP】

最新推荐文章于 2018-09-25 22:41:45 发布

AC_Dreameng

最新推荐文章于 2018-09-25 22:41:45 发布

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分类专栏： ZOJ ACM_HDU刷题录文章标签： POJ1050 To the Max DP

本文链接：https://blog.youkuaiyun.com/hurmishine/article/details/70187638

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这篇博客详细介绍了如何解决POJ 1050 - To the Max的问题，重点在于使用动态规划（DP）方法。作者建议在尝试此题之前先理解HDU 1003，并将二维问题转化为一维。博客提供了问题的输入、输出格式，示例输入和输出，以及AC（Accepted）代码。

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To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48554		Accepted: 25678

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

Source

Greater New York 2001

原题链接：http://poj.org/problem?id=1050

做这题前，先弄清楚HDU1003这题。

然后再把二维压缩为一维就可以了。

1,12,123,1234，2,23,234,3,34,4

按上面几种情况压缩，求出最大的即可。

AC代码：

/**
  * 行有余力，则来刷题！
  * 博客链接:http://blog.youkuaiyun.com/hurmishine
  *
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=100+5;
int a[maxn][maxn];
int dp[maxn];
int n;
int getMax(int *a)
{
    int maxx=a[0];
    int sum=a[0];
    for(int i=1;i<n;i++)
    {
        if(sum+a[i]>a[i])
            sum+=a[i];
        else
            sum=a[i];
        if(sum>maxx)
            maxx=sum;
    }
    return maxx;
}
int main()
{
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
                cin>>a[i][j];
        }
        int p;
        int maxx=0;
        for(int i=0;i<n;i++)
        {
            memset(dp,0,sizeof(dp));
            for(int j=i;j<n;j++)
            {
                p=0;
                for(int k=0;k<n;k++)
                {
                    dp[p]+=a[j][k];
                    p++;
                }
                int ans=getMax(dp);
                if(ans>maxx)
                    maxx=ans;
            }
        }
        cout<<maxx<<endl;
    }
    return 0;
}