ZOJ1074-To the Max（dp）

To the Max

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 

思路：将二维最大子矩阵问题转化为一维最大连续子串和问题；一维最大连续子串和：点击打开链接

只需通过压缩行，将其变为一维就可用一维最大连续子串和的方式去解决；

代码：

#include<stdio.h>#include<string.h>#define N 101int main(){int a[N][N];int b[N];int n;int i,j,k;while(scanf("%d",&n)!=EOF){for(i=0; i<n; i++)for(j=0; j<n; j++)scanf("%d",&a[i][j]);int max = -32767;for(i=0; i<n; i++){memset(b, 0, sizeof(b));for(j=i; j<n; j++){int sum=0;for(k=0; k<n; k++){b[k] += a[j][k];sum += b[k];if(sum<0) sum = b[k];if(sum>max) max = sum;}}}printf("%d\n",max);}return 0;}