HDU 1081 最大子矩阵和

Posted 2020-08-09 半根毛线code

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：

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 //#include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cmath>
11 #include<cstdio>
12 #define ll long long
13 #define mod 1000000007
14 #define PI acos(-1.0)
15 using namespace std;
16 int main()
17 {
18     int a[105][105];
19     int b[105];
20     int n;
21     while(scanf("%d",&n)!=EOF)
22     {
23         for(int i=1;i<=n;i++)
24         {
25             for(int j=1;j<=n;j++)
26                 scanf("%d",&a[i][j]);
27         }
28         int maxn=-100000;
29         for(int i=1;i<=n;i++)
30         {
31             memset(b,0,sizeof(b));
32             for(int j=i;j<=n;j++)
33             {
34                 int sum=0;
35                 for(int k=1;k<=n;k++)
36                 {
37                     b[k]+=a[j][k];
38                     sum+=b[k];
39                     if(sum<0) sum=b[k];
40                     if(sum>maxn) maxn=sum;
41                 }
42             }
43         }
44         printf("%d\\n",maxn);
45     }
46     return 0;
47 }