hdu - To The Max-1081 - 最大子矩阵 - 最大连续子序列和变形/动态规划

问题

Problem 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

15

题解

思路

这是一道求子矩阵的最大和问题，由于数据给得比较小，n≤100，可以使用常规的O(N^3)的时间复杂度的算法解，枚举每一行以及该行下面的所有的行的和，时间复杂度为O(N^2)，然后对压缩后的每一行进行dp，压缩完后的tem[i]就对应一维数组的连续子序列和问题，时间复杂度为线性的O(N)，最后总的时间复杂度为O(N^3)

一般的矩阵是二维的，我们要把它压缩为一维的。

即将一层层的数相加，压缩成一维的即可，我们遍历所有的压缩可能，就可以求子序列和的方法求出最大值。

dp的状态转移方程：dp[i] = MAX(dp[i-1] + tmp[i], dp[i])

AC代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long LL;

int a[105][105];
int dp[105],tmp[105];
int main()
{
int n;
while(~scanf("%d",&n)){
for(int i=0;i<n;++i)
for(int j=0;j<n;++j)
scanf("%d",&a[i][j]);
int maxn=-1e7;
for(int i=0;i<n;i++){枚举每一行
memset(tmp,0,sizeof(tmp));
for(int t=i;t<n;t++){//求i及以下小矩阵最大值
for(int j=0;j<n;j++)
tmp[j]+=a[t][j];
dp[0]=tmp[0];
for(int j=1;j<n;j++){
if(dp[j-1]>0)
dp[j]=dp[j-1]+tmp[j];
else
dp[j]=tmp[j];
if(dp[j]>maxn) maxn=dp[j];
}
}
}
printf("%d\n",maxn);
}
return 0;
}