Maximum Sum

Maximum Sum

 

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem. Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the subrectangle with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array. 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-hand corner: 9 2 −4 1 −1 8 and has the 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 N2 integers separated by white-space (newlines and spaces). These N2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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 The output is 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

 

题意：给一个N*N大小的矩阵，每个格子上有一个数字。 求这个矩阵中的子矩阵上面数字之和最小的数字是多少

思路: 把矩阵缩小，上下两个数变为一个数，缩小的矩阵变成若干个小矩阵，比较小矩阵中和比较大的那个，在积累矩阵的和时，当为负数时直接放弃

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
int map[110][110];
int main()
{
	int n;
	while(~scanf("%d",&n)&&n)
	{
		int i,j;
		for(i=0;i<n;i++)
		 for(j=0;j<n;j++)
		 scanf("%d",&map[i][j]);
		int maxx=map[0][0];
		for(i=1;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				map[i][j]+=map[i-1][j];//把上下两个数之和当成一个数 
			}
		 } 
		 for(i=0;i<n;i++)
		 {
		 	for(j=i;j<n;j++)
		 	{
		 		int sum=0;
		 		for(int k=0;k<n;k++)
		 		{
		 			if(sum<0)//当出现小于0的和直接放弃 
		 			   sum=0;
		 			else if(i!=j)
		 			  sum+=map[j][k]-map[i][k];//矩阵的正数之和 
		 			if(sum>maxx&&sum>0)//比较正数中比较大的那个 
		 			  maxx=sum;
				 }
			 }
		 }
		 printf("%d\n",maxx);
	}
}