ZOJ1074 最大和子矩阵 DP

To the Max

Time Limit: 1 Second      Memory Limit: 32768 KB

Problem

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.

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.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15

 

 

DP 求最大和子矩阵 , 可以用最大和连续子序列的思路解 .

首先 , 读数据的时候 a[i][j] 为前 i 行数据的第 j 个数的和 , 当求第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵的时候 , 就以 p 行和 q 行之间的同一列的数字的和作为一个数列的元素 , 然后 DP 求这个数列的最大和连续子序列 , 就是第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵 .

代码如下 :

 

 

#include<stdio.h> #include<string.h> #define MAXN 105 int main() { int i,j,k,n,t,sum,max; int a[MAXN][MAXN]; while (scanf("%d",&n)!=EOF) { memset(a,0,sizeof(a)); for (i=1;i<=n;++i) { for (j=1;j<=n;++j) { scanf("%d",&t); a[i][j]=a[i-1][j]+t; } } max=0; for (i=1;i<=n;++i) { for (j=i;j<=n;++j) { sum=0; for (k=1;k<=n;++k) { t=a[j][k]-a[i-1][k]; sum+=t; if (sum<0) sum=0; if (sum>max) max=sum; } } } printf("%d/n",max); } return 0; }