POJ 1050-To the Max（最大子矩阵和）

女爷i 2022-06-17 11:36 130阅读 0赞

To the Max

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;48640</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;25719</td> 
  </tr> 
 </tbody> 
</table>

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][]

## 题目意思： ##

N\*N矩阵，求其中一个最大的子矩阵使得其和最大。

## 解题思路： ##

转换成一维DP的思路来做，先处理前缀和，对于每一列上的元素，当前元素的值加上其前面所有元素的值。

然后参考一维DP的做法，枚举起始和终止行，在列上找最大值。

#include<iostream>
    #include<cstdio>
    #include<iomanip>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<map>
    #include<algorithm>
    #include<vector>
    #include<queue>
    using namespace std;
    #define INF 0xfffffff
    #define MAXN 101
    int a[MAXN][MAXN];
    int b[MAXN];
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        int n;
        cin>>n;
        memset(a,0,sizeof(a));
        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];//列上的元素依次求和
            }
        /*for(int i=1; i<=n; ++i)
        {
            for(int j=1; j<=n; ++j)
                cout<<a[i][j]<<" ";
            cout<<endl;
        }*/
        int ans=a[1][1];
        for(int i=0; i<=n-1; ++i)//起始行
            for(int j=i+1; j<=n; ++j)//终止行
            {
                memset(b,0,sizeof(b));//第i~j行k列中能取得的最大值
                for(int k=1; k<=n; ++k)//当前枚举的列
                {
                    if(b[k-1]>=0) b[k]=b[k-1]+a[j][k]-a[i][k];
                    else b[k]=a[j][k]-a[i][k];
                    ans=max(ans,b[k]);
                }
            }
        cout<<ans<<endl;
    }

[Greater New York 2001]: http://poj.org/searchproblem?field=source&key=Greater+New+York+2001