动态规划之To the Max

二维数组最大子矩阵和

最新推荐文章于 2021-09-05 08:59:11 发布

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最新推荐文章于 2021-09-05 08:59:11 发布

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分类专栏： ACM

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本文介绍了一种解决二维数组中寻找最大子矩阵和的方法，通过将二维问题转化为一维的最大子段和问题来简化计算过程，并给出了具体的实现代码。

题目：

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

这道题一开始想歪了....后来看了别人的，发现只是把一个二维的压缩成一维的，然后问题就转化为求最大子段和的问题了

代码如下（代码写的好捉急的说...）：

#include<stdio.h>
int num[100][100];
int sum[100];
int f[100];
int N;
int sum_col(int i,int j,int col){
    int ans,row;
    ans=0;
    for(row=i;row<j+1;row++)
        ans+=num[row][col];
    return ans;
}
int main(){
    int i,j,k,col,ans,tmp;
    tmp=ans=-1300000;
    scanf("%d",&N);
    for(i=0;i<N;i++)
        for(j=0;j<N;j++)
            scanf("%d",&num[i][j]);
    for(i=0;i<N;i++)
        for(j=i;j<N;j++){
            for(col=0;col<N;col++)
                sum[col]=sum_col(i,j,col);
            for(k=0;k<N;k++){
                if(f[k-1]<0)
                    f[k]=sum[k];
                else
                    f[k]=sum[k]+f[k-1];
                if(f[k]>tmp)
                    tmp=f[k];
            }
            if(tmp>ans)
                ans=tmp;
        }
    printf("%d\n",ans);
    return 0;
}