To The Max_hdu_1081(经典DP).java

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6815    Accepted Submission(s): 3270


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

 

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import java.io.InputStreamReader;
import java.util.Scanner;

public class Main{//经典dp
public static void main(String[] args) {
Scanner input=new Scanner(new InputStreamReader(System.in));
while(input.hasNext()){
int n=input.nextInt();
int a[][]=new int[n+1][n+1];
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
a[i][j]=a[i][j-1]+input.nextInt(); //表示第i行前j个元素之和
}
}
int max=-999;
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
int sum=0;
for(int k=1;k<=n;k++){
if(sum<0)
sum=0;
sum+=a[k][j]-a[k][i-1];//计算第k行第j和第i列之间的的数
if(sum>max)
max=sum;
}
}
}
System.out.println(max);
}
}
}

#include"stdio.h"
#include"string.h"
int main()
{
int num[105][105];
int i1,i2,i_temp,l;
int temp[105];
int n;
int max;
int temp0;
while(scanf("%d",&n)!=-1)
{
for(i1=0;i1<n;i1++)
for(l=0;l<n;l++)
scanf("%d",&num[i1][l]);

max=0;
for(i2=0;i2<n;i2++)
{
for(i1=0;i1<=i2;i1++)
{
/*****/                  //压缩
for(l=0;l<n;l++)
{
temp0=0;
for(i_temp=i1;i_temp<=i2;i_temp++)	temp0+=num[i_temp][l];
temp[l]=temp0;
}
/*****/

if(temp[0]>max)	max=temp[0];
for(l=1;l<n;l++)
{
if(temp[l-1]>0)	temp[l]+=temp[l-1];
if(temp[l]>max)	max=temp[l];
}
}
}

printf("%d\n",max);
}
return 0;
}

#include<iostream>
using namespace std;
int a[101][101];

int main()
{
int n;
int i,j,k;
int temp;
while(cin>>n)
{
memset(a,0,sizeof(a));
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
cin>>temp;
a[i][j]+=a[i][j-1]+temp;           //表示第i行前j个元素之和
}
int max=-999,sum;

for(i=1;i<=n;i++)
{
for(j=i;j<=n;j++)
{
sum=0;
for(k=1;k<=n;k++)
{
if(sum<0)
sum=0;

sum+=a[k][j]-a[k][i-1];   //计算第k行第j和第i列之间的的数

if(sum>max)
max=sum;
}
}
}

printf("%d\n",max);

}
return 0;
}
/*
4
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
*/