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
给定一个N*N的矩阵，要求一个子矩阵，使得该矩阵的元素和最大。转为为最大连续子序列问题。
1 #include <iostream>
2
3 const int N = 101;
4 const int inf = INT_MIN;
5 int a[N][N],b[N];
6
7 int MaxSum(int n,int a[]){
8 int i,sum=inf,b=0;
9 for(i=1;i<=n;i++){
10 b=(b+a[i])>a[i]?(b+a[i]):a[i];
11 sum=sum>b?sum:b;
12 }
13 return sum;
14 }
15 int main(){
16 int n,i,j,k,max,ans;
17 while(scanf("%d",&n)!=EOF){
18 for(i=1;i<=n;i++)
19 for(j=1;j<=n;j++)
20 scanf("%d",&a[i][j]);
21 for(max=inf,i=1;i<=n;i++){
22 memset(b,0,sizeof(b));
23 for(j=i;j<=n;j++){
24 for(k=1;k<=n;k++)
25 b[k]+=a[j][k];
26 ans=MaxSum(n,b);
27 max=max>ans?max:ans;
28 }
29 }
30 printf("%d\n",max);
31 }
32 return 0;
33 }