hdu 1081 to be max

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

 

算法：最小子列和比较简单，二维问题可以通过枚举一个i,j降为一维问题。

代码：

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
using namespace std;

int dp[110];

int m[110][110];
int PreSum[110][110];
int ary[110];
int n;

int Result;

bool Input();
void Caculate();
int GetDp();

int main(){
	while(cin>>n){
		Input();
		Caculate();
		cout<<Result<<endl;
	}
}

bool Input(){
	memset(PreSum, 0, sizeof(PreSum));
	for(int i = 0; i < n; ++i){
		for(int j = 1; j <= n; ++j){
			cin>>m[i][j];
			PreSum[i][j] = PreSum[i][j-1] + m[i][j];
		}
	}
	return true;
}

void Caculate(){
	Result = -0x0FFFFFFF;
	for(int i =0; i <= n; ++i){
		for(int j = i + 1; j <= n; ++j){
			for(int l = 0; l < n; ++l){
				ary[l] = PreSum[l][j] - PreSum[l][i];
			}
			Result = max(Result, GetDp());
		}
	}
}

int GetDp(){
	memset(dp, 0, sizeof(dp));
	dp[0] = ary[0];
	for(int i = 1; i < n; ++i){
		dp[i] = max(dp[i-1] + ary[i], ary[i]);
	}
	return *max_element(dp, dp + n);
}

结果：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
15822302	2015-12-11 21:45:43	Accepted	1081	31MS	1888K	963 B	C++	BossJue