动态规划——F 最大矩阵和
2015-08-13 22:14
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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
Sample Output
View Code
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 解题思路: 最大子矩阵,首先一行数列很简单求最大的子和,我们要把矩阵转化成一行数列, 就是从上向下在输入的时候取和,a[i][j]表示在J列从上向下的数和,这样就把一列转化成了一个点, 再用双重,循环,任意i行j列开始的一排数的最大和,就是最终的最大和 注意:如果m<0,则就不需要继续加了 程序代码:
#include <cstdio> #include <cstring> using namespace std; int a[110][110]; int main() { int n,c; while( scanf("%d",&n)==1) { memset(a,0,sizeof(a)); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { scanf("%d",&c); a[i][j]=a[i-1][j]+c; } int sum=0; for(int i=1;i<=n;i++) for(int j=i;j<=n;j++) { int m=0; for(int k=1;k<=n;k++) { int t=a[j][k]-a[i-1][k]; m+=t; if(m<0) m=0; if(sum<m) sum=m; } } printf("%d\n",sum); } return 0; }
View Code
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