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.
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
思路:
这题就是求最大子矩阵,可以先把每行的前缀和求出来,然后枚举区间长度,然后列向进行最大区间和dp,总共需要运算 ∑n*(n-i)(1<=i<=n)次,复杂度o(n^3)。下面是代码:
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
typedef long double ld;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const int MAXN=101;
int sum[MAXN][MAXN];
int DP(int l,int n)//区间长度为l的dp
{ int maxx=-INF,dp=0;
for(int i=0;i<=n-l;i++)
{ dp=0;
for(int j=1;j<=n;j++)
{if(dp>=0) dp=dp+sum[j][i+l]-sum[j][i];
else dp=sum[j][i+l]-sum[j][i];
if(dp>maxx) maxx=dp;
}
}
return maxx;
}
int main()
{ int ans;
int n,x;
while(~scanf("%d",&n))
{ ans=-INF;
memset(sum,0,sizeof(sum));
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{scanf("%d",&x);
sum[i][j]=sum[i][j-1]+x;
}
for(int i=1;i<=n;i++) ans=max(ans,DP(i,n));
printf("%d\n",ans);
}
}