Problem Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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:题目不难理解呀。就是给你一个nn的矩阵,让你找出和最大的子矩阵来。
这种题原来做过,好久没看了,忘了。。。。大体的思路就是先将原来矩阵每一行的数存到一个nn的前缀和数组里。然后控制列数,移动行数,这样来找寻最大子矩阵。有的题解说这样是二维压缩成一维,我也不太明白。反正思路明白就好了。
代码如下:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int a[101][101];
int n;
int temp;
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&temp);
a[i][j]=a[i][j-1]+temp;
}
}
int ans=-inf;
int res=0;
for(int i=1;i<=n;i++)
{
for(int j=i;j<=n;j++)
{
for(int k=1,res=0;k<=n;k++)//每次行数从头开始的时候就要将res更新为0
{
res+=(a[k][j]-a[k][i-1]);
if(res<0) res=0;//res小于0时,res就要更新为零!!!
ans=max(ans,res);
}
}
}
printf("%d\n",ans);
}
return 0;
}
努力加油a啊,(o)/~