Background
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.
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
sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size  or
greater located within the whole array. As an example, the maximal sub-rectangle of the array:
is in the lower-left-hand corner:
and has the sum of 15.
Input and Output
The input consists of an  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  integers
separated by white-space (newlines and spaces). These 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].
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:给定一个矩阵。要求出一个和最大的子矩阵之和。。
思路:和一维的求最大和子序列有点类似。不过这是二维的。直接暴力会超时。所以每次进行操作的时候。要把值保留下来。以备给后面使用。这样可以大大减少所用的时间。。
代码:
#include <stdio.h>
#include <string.h>
#include <limits.h>
int n, num[105][105], Max, he, sum[105];
void init() {
Max = -INT_MAX;
for (int i = 0; i < n ; i ++)
for (int j = 0 ; j < n ; j ++)
scanf("%d", &num[i][j]);
}
int solve() {
for (int i = 0; i < n ; i ++) {
memset(sum, 0, sizeof(sum));
for (int j = i; j < n; j ++) {
he = 0;
for (int k = 0; k < n; k ++) {
sum[k] += num[j][k];
if (he >= 0)//这步跟求最大子序列和一个原理。。如果加到是负数的时候，是会使总和减少的。所以直接从下一个位置开始作为起点。
he += sum[k];
else
he = sum[k];
if (Max < he)
Max = he;
}
}
}
return Max;
}
int main() {
while (~scanf("%d", &n)) {
init();
printf("%d\n", solve());
}
return 0;
}
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