To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 39913 | Accepted: 21099 |
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:
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].
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
Source
这是hdu 1003的拓展，知道了在一维数组中如何求最大连续子段和，那么这题就是扩展到二维数组中，让我们求出子矩阵最大的和，我们可以这样考虑，我们把同行不同列（或者同列不同行）的加起来，比如i行，j行，i,j两行之间的数字组成了一个矩阵，我们把i行到j行之间同列的数组元素加起来按照列号组成一个一维数组，这样我们只需要利用最大子段和的知识找出这个数组的最大连续和，这个和就是我们要求的那个子矩阵最大和！ 可以说， i和j行定义了子矩阵高度，一维数组最大子段和连续的长度定义了子矩阵的宽度，OK，代码。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std; int temp[],n;
int cal() //
{
int Max = temp[];
int sum = temp[];
for(int i = ;i<n;i++)
{
if(sum+temp[i]<temp[i])
sum = temp[i];
else
sum+=temp[i];
if(Max<sum)
Max = sum;
}
return Max;
} int main()
{
while(scanf("%d",&n)!=EOF)
{
int a[][];
for(int i = ;i<n;i++)
for(int j = ;j<n;j++)
scanf("%d",&a[i][j]);
int Max = ;
for(int i = ;i< n;i++) //i是起始行
{
for(int j =i ;j<n;j++) //j是终止行
{
memset(temp,,sizeof(temp));
for(int m = ;m<n;m++) //固定列,注意是行在变
{
for(int k =i ;k<=j;k++) //累加i起始行，j终止行中间的同列的数据
temp[m]+=a[k][m];
}
int MaxTemp = cal();
if(MaxTemp>Max)
Max = MaxTemp;
} }
printf("%d\n",Max);
}
return ;
}
事实证明我蠢了，后来看到nyoj这题的最优程序解答，在处理第i行到j行同列相加上面处理的很好，利用输入时候进行累加，然后做减法，直接减掉了一层循环
nyoj 的版本
#include<iostream>
#include<cstring>
using namespace std;
#define N 110
int a[N][N];
int b[N];
int main()
{
int n,r,c;
cin>>n;
while(n--)
{
cin>>r>>c;
for(int i=;i<=r;++i)
for(int j=;j<=c;++j)
{
cin>>a[i][j];
a[i][j]+=a[i-][j];
}
int max=a[][];
for(int i=;i<=r-;++i)
for(int j=i+;j<=r;++j)
{
memset(b,,sizeof(b));
for(int k=;k<=c;++k)
{
if(b[k-]>=)
b[k]=b[k-]+a[j][k]-a[i][k];
else
b[k]=a[j][k]-a[i][k];
if(max<b[k])
max=b[k];
}
}
cout<<max<<endl;
}
}
还有一种没有用这种求和的方法，但是是先求第一行最大子段和，再求第一行跟第二行合起来的最大子段和，，再求第一到第三合起来的最大子段和，以此类推，直到求出整个矩阵的合起来的最大子段和，最后就是我们需要的解 。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std; int temp[],n;
int cal() //最大子段和
{
int Max = temp[];
int sum = temp[];
for(int i = ;i<n;i++)
{
if(sum+temp[i]<temp[i])
sum = temp[i];
else
sum+=temp[i];
if(Max<sum)
Max = sum;
}
return Max;
} int main()
{
while(scanf("%d",&n)!=EOF)
{
int a[][];
for(int i = ;i<n;i++)
for(int j = ;j<n;j++)
scanf("%d",&a[i][j]);
int Max = ;
for(int i = ;i< n;i++)
{
memset(temp,,sizeof(temp));
for(int j =i ;j<n;j++)
{
for(int k = ;k<n;k++)
temp[k]+=a[j][k];
int t = cal();
if(t>Max)
Max =t;
} }
printf("%d\n",Max);
}
return ;
}