【dp 最大子矩形】poj1050
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.
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
最大子矩形的方法是先压维，即将几行压到一行里，然后最大子序列和就可以了（把几行压到一行里的方法是先预处理个前缀和）
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const long INF=2^32;
long n;
long a[150][150];
long sumy[150][150];
long f[150][150][150];
void dp()
{
for(long i = 1; i <= n; i++)
for(long j = 1; j <= n; j++)
{
sumy[i][j] = sumy[i - 1][j] + a[i][j];
}
for(long k = 1; k <= n; k++)
for(long i = k; i <= n; i++)
for(long j = 1; j <= n; j++)
{
f[k][i][j] = max(sumy[i][j] - sumy[k - 1][j], sumy[i][j] - sumy[k - 1][j] + f[k][i][j - 1]);
}
long maxn = -INF;
for(long k = 1; k <= n; k++)
for(long i = k; i <= n; i++)
for(long j = 1; j <= n; j++)
{
if (f[k][i][j] > maxn) maxn = f[k][i][j];
}
cout << maxn << endl;
}
int main()
{
while (cin >> n)
{
for (long i = 1; i <= n; i++)
{
for (long j = 1; j <= n; j++)
{
scanf("%d", &a[i][j]);
for (long k = 1; k <= n; k++)
f[i][j][k] = -INF;
}
}
memset(sumy, 0, sizeof(sumy));
dp();
}
return 0;
}
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