POJ- 1050- to the Max 【动态规划】

To the Max

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

题意:题目的意思很简单，就是求已知的二维数组中最大的子二维数组和（说子矩阵不合适，就说成子数组，理解就行）

题解：我们都处理过一维的，现在是二维的，所以我们要想办法把二维的转化成一维处理。我们用一个二维数组记录每一行前j列的和，然后用sum[k][j]-sun[k][i-1]，就说明第k行第i列到第j列的和。具体见代码：

AC：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<climits>
using namespace std;
const int maxn=101;
int arr[maxn][maxn]={0},sum[maxn][maxn]={0};
int main(void){
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
cin>>arr[i][j];
sum[i][j]=sum[i][j-1]+arr[i][j]; //记录每一行前j列的和
}
}
int max=INT_MIN,temp;//max初始化为整数最小值
for(int i=1;i<=n;i++){ //列
for(int j=i;j<=n;j++){ //列
temp=0;
for(int k=0;k<=n;k++){ //行
if(temp<0) temp=0;
temp+=sum[k][j]-sum[k][i-1];
if(temp>max) max=temp; //一旦有比之前的最大值就取代记录
}
}
}
printf("%dn",max);
return 0;
}

最后

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