To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10874 Accepted Submission(s): 5204

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

动态规划问题

可以将二维数组压缩成一维数组，求压缩后的数字最大子段和，即二维数组的最大子矩阵和。

将二维数组压缩：将第1行到第n行的每一列都加起来，即a[i][j]表示第j列前1-i行数的和，再求第i行最大子段和就可以了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 int main(){
 7     int n, a[110][110], i, j, k;
 8     while(cin>>n){
 9         memset(a, 0, sizeof(a));
10         for(i=1; i<=n; i++){
11             for(j=1; j<=n; j++){
12                 cin>>a[i][j];
13                 a[i][j] += a[i-1][j];//a[i][j]表示第j列前i行数的和
14             }
15         }
16         int mmax = -1000000, sum;
17         for(i=1; i<=n; i++){
18             for(k=1; k<=i; k++){
19                 sum = 0;
20                 for(j=1; j<=n; j++){
21                     sum += (a[i][j]-a[k-1][j]);
22                     if(sum > mmax)
23                         mmax = sum;
24                     if(sum < 0)
25                         sum = 0;
26                 }
27             }
28         }
29         cout<<mmax<<endl;
30     }
31     return 0;
32 }