poj 1050最大子矩阵

最新推荐文章于 2021-04-08 08:41:24 发布

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本文介绍了一种求解二维数组中最大子矩阵和的问题，并提供了一个具体的算法实现。该算法通过对数组进行预处理，利用前缀和思想，将问题转化为一维最大子数组和问题进行求解。

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题目：

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49846		Accepted: 26420

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

分析：

不需要二维树状数组因为没有修改操作只要维护一个前缀和就可以了

具体参：http://blog.pureisle.net/archives/266.html 讲得很好

代码：

#include<iostream>
#include<cstring>
using namespace std;
#define N 110
int a[N][N],f[N];
int main(){
    int n,r;
    cin>>r;
    for(int i=1;i<=r;++i){
        for(int j=1;j<=r;++j){
            cin>>a[i][j];
            a[i][j]+=a[i-1][j];///a[i][j]代表前i行 第j列的部分和
        }
    }
    int ans=a[1][1];
    for(int i=0;i<=r-1;++i){///起始行
        for(int j=i+1;j<=r;++j){///终止行
            ///相当于每次对[i+1,j]行进行操作，将这几行捆绑在一起进行考虑
            ///a[j][k]-a[i][k]即为[i+1,j]行第k列的部分和
            memset(f,0,sizeof(f));
            for(int k=1;k<=r;++k){///转换为一维最大子串和 a[j][k]-a[i][k]相当于一维数组里面每次判断的量
                if(f[k-1]>=0) f[k]=f[k-1]+a[j][k]-a[i][k];
                else f[k]=a[j][k]-a[i][k];
                if(ans<f[k]) ans=f[k];
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}