To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 45915 | Accepted: 24282 |
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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
1 /*这道题目n^3居然不会超时，暴力就好了*/
2 #include<iostream>
3 using namespace std;
4 #include<cstdio>
5 #define N 101
6 int sum[N][N]={0},n,x;
7 int main()
8 {
9 scanf("%d",&n);
10 for(int i=1;i<=n;++i)
11 for(int j=1;j<=n;++j)
12 {
13 scanf("%d",&x);
14 sum[i][j]=sum[i][j-1]+x;/*sum[i][j]表示的是第i行前j个数的和*/
15 }
16 int ans=-(1<<30);
17 for(int i=1;i<=n;++i)
18 for(int j=i;j<=n;++j)/*暴力枚举每行区间*/
19 {
20 int tmp=0;/*tmp为当前矩阵的大小*/
21 for(int k=1;k<=n;++k)/*枚举每一行*/
22 {
23 int que=sum[k][j]-sum[k][i-1];/*取出这一行*/
24 if(tmp>0) tmp+=que;/*如果当前矩阵的大小已经<0了，那么再加上就要放弃之前的矩阵，放弃一定会更优*/
25 else tmp=que;
26 ans=max(ans,tmp);/*因为我们会随时放弃矩阵，所以最大值的更新，在循环中进行*/
27 }
28 }
29 cout<<ans;
30 return 0;
31 }