hdu 1081 To The Max
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4825 Accepted Submission(s): 2281
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.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define N 133
using namespace std;
int dpsum[N][N];
int a[N][N];
int Max,n;
void dp(int &i,int &len)
{
int r=i+len-1,l=i-1;
int k,m_ax,b=dpsum[1][r]-dpsum[1][l];
m_ax=b;
for(k=2;k<=n;k++)
{
if(b<0)
b=dpsum[k][r]-dpsum[k][l];
else
b+=dpsum[k][r]-dpsum[k][l];
if(b>m_ax)
m_ax=b;
}
Max=max(Max,m_ax);
}
int main()
{
int i,j,k;
int l;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
scanf("%d",&a[i][j]);
dpsum[i][j]=dpsum[i][j-1]+a[i][j];
}
Max=0;
for(l=1;l<=n;l++)//求宽度为l的子矩阵和
{
j=n-l+1;
for(i=1;i<=j;i++)
dp(i,l);
}
printf("%d\n",Max);
}
return 0;
}