HDU 1081 最大子矩阵和

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11715    Accepted Submission(s): 5661

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

 

题意：给你一个n*n的矩阵  输入比较恶心 无视就可以了 问最大的子矩阵和

题解：

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#define ll long long
#define mod 1000000007
#define PI acos(-1.0)
using namespace std;
int main()
{
    int a[105][105];
    int b[105];
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                scanf("%d",&a[i][j]);
        }
        int maxn=-100000;
        for(int i=1;i<=n;i++)
        {
            memset(b,0,sizeof(b));
            for(int j=i;j<=n;j++)
            {
                int sum=0;
                for(int k=1;k<=n;k++)
                {
                    b[k]+=a[j][k];
                    sum+=b[k];
                    if(sum<0) sum=b[k];
                    if(sum>maxn) maxn=sum;
                }
            }
        }
        printf("%d
",maxn);
    }
    return 0;
}