To The Max
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
就是给一个n*n的矩阵，求出它的最大子矩阵
我们都做过求一个序列的最大子序列吧。O(n)的复杂度。
这道题就是那道题转化一下。
暴力枚举第k1行到第k2行每一列各个元素的和sum[i]，然后对sum[i]进行一次最大子序列求和
然后不断更新答案
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4
5 using namespace std;
6
7 const int maxn=105;
8
9 int a[maxn][maxn];
10 int sum[maxn];
11 int dp[maxn];
12
13 int main()
14 {
15 int n;
16 while(scanf("%d",&n)!=EOF)
17 {
18 for(int i=1;i<=n;i++)
19 for(int j=1;j<=n;j++)
20 scanf("%d",&a[i][j]);
21
22 int ans=-1000000;
23
24 for(int i=1;i<=n;i++)
25 {
26 memset(dp,0,sizeof(dp));
27 memset(sum,0,sizeof(sum));
28 for(int j=i;j<=n;j++)
29 {
30 for(int k=1;k<=n;k++)
31 {
32 sum[k]+=a[j][k];
33 }
34
35 for(int k=1;k<=n;k++)
36 {
37 dp[k]=max(dp[k-1],0)+sum[k];
38 if(dp[k]>ans)
39 ans=dp[k];
40 }
41
42 }
43 }
44
45 printf("%d\n",ans);
46 }
47
48 return 0;
49 }