最大子矩阵面积

参考题目 zoj 1074

所需知识 求最大子串和

ZOJ Problem Set - 1074
To the Max

Time Limit: 2 Seconds Memory Limit: 65536 KB

Problem

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.

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.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15

/*
用二维数组sum来把最大子矩阵转化为最大子串和问题即可 
因为对n行的矩阵，子矩阵肯定是1~n行 所以办法就是把所有行看出一个单位进行纵放心取子串
对于每个行子串展开看即为一个矩阵，把矩阵上下相加，变成了一个横向的数组，进行求最大子串和 
(eg:3行m列，则1，2，3行单独求一遍最大子串，第一行加上第二行求一遍，2+3，1+3求一遍，然后1+2+3求一遍，取最大的即可) 
*/

#include<iostream>
#include<cstdio>
#define N 1001
#include<cstring>
#define sub(i,a,b) sum[b][i]-sum[a-1][i]
using namespace std;
int n;
int M[N][N];
int sum[N][N];//sum[i][j]表示 
int ans=-0x7fffffff;
int dp;

void IN(void){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){//养成好习惯从1开始，很多时候需要用到0 
        for(int j=1;j<=n;j++){
            scanf("%d",&M[i][j]);
            sum[i][j] = sum[i-1][j]+M[i][j];//前缀和 
        }
    }
}

void OUT(void){
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){//从i行到j行
            //printf("case %d to %d\n",i,j); 

            for(int k=1;k<=n;k++){//k 从左往右 
                if(k==1)dp=sub(k,i,j);
                else dp=max(dp+sub(k,i,j),sub(k,i,j));

                if(dp>ans){
                    ans=dp;//printf("from line %d to %d , when k is %d , update \n",i,j,k);
                    //printf("ans is to %d\n",ans);
                }
            }
        }
    }
    cout<<ans<<endl; 
}

int main()
{
    IN();
    OUT();
    return 0;
}
/*
4
 0 -2 -7  0
 9  2 -6  2
-4  1 -4  1
-1  8  0 -2

out 15
*/