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.
4 0 -2 -7
0 9 2 -6 2
-4 1 -4 1
-1 8 0 -2
Sample Output
15
题目大意：找到和是最大的钜形；
解题思路：和一维的差不多，不过是把二维的变成一维的然后再找到最大的和。
代码如下：
#include<stdio.h>
#include<string.h>
int main()
{
int n;
int mp[200][200],v[500];
while(~scanf("%d",&n))
{
int i,j,k;
for(i=0; i<n; i++)
for(j=0; j<n; j++)
scanf("%d",&mp[i][j]);
int maxx=-101100;
for(i=0; i<n; i++)
{
memset(v,0,sizeof(v));
for(j=i; j<n; j++)
{
int sum1=0;
for(k=0; k<n; k++)
{
v[k]+=mp[j][k];//计算横行的和，
if(sum1<0)
sum1=v[k];
else
sum1=sum1+v[k];
if(sum1>maxx)
maxx=sum1;
}
}
}
printf("%d\n",maxx);
}
return 0;
}