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): 4437    Accepted Submission(s): 2099

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
   
   
   
   

Source

Greater New York 2001

 

可以根据最大子段，引申出来

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 50008
using namespace std;
int prim[N],numprim[N];
bool isnoprime[N];
int next;
void primm()
{
  memset(isnoprime,0,sizeof(isnoprime));
  next=0;
  int i,j,k;
  for(i=2;i<N;i++)
  {
    if(!isnoprime[i])
    {prim[++next]=i;}
    for(j=i+i;j<N;j++)
    {
      isnoprime[j]=1;
    }
  }
}
void DP()
{
  memset(numprim,999999,sizeof(numprim));
  numprim[0]=0;
  numprim[1]=0;
  int k;
  for(int i=0;i<N;i++)
  {
    for(int j=1;j<=next&&(i+prim[j])<=N;j++)
    {
      numprim[i+prim[j]]=min(numprim[i+prim[j]],numprim[i]+1);

    }
    for(int j=1;j<=next&&i*prim[j]<=N;j++)
    numprim[i*prim[j]]=min(numprim[i*prim[j]],numprim[i]+1);
  }
}
int main()
{
  int m;
  scanf("%d",&m);
  primm();
  DP();
  int a,b;
  while(m--)
  {
    scanf("%d%d",&a,&b);
    int sum=0;
    for(int i=a;i<=b;i++)
    sum+=numprim[i];
    printf("%d\n",sum);
  }
}