Description
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:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <map>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define MAXX 100+10
#define inf 0x3f3f3f3f
using namespace std;
int maxsum(int *a, int n)
{
int b=0, sum=-2147483645;
for (int i=0; i<n; i++)
{
if (b<0)
b = a[i];
else
b += a[i];
if (b>sum)
sum = b;
}
return sum;
}
int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // LOCAL
int n;
while (scanf("%d", &n)!=EOF)
{
int a[MAXX][MAXX], b[MAXX];
for (int i=0; i<n; i++)
for (int j=0; j<n; j++)
scanf("%d", &a[i][j]);
int ans = -2147483645;
for (int i=0; i<n; i++)
{
memset(b, 0, sizeof(b));
for (int j=i; j<n; j++)
{
for (int k=0; k<n; k++)
b[k] += a[j][k];
int maxx = maxsum(b, n);
if (maxx > ans)
ans = maxx;
}
}
printf("%d\n", ans);
}
return 0;
}
相关的知识: http://blog.youkuaiyun.com/liufeng_king/article/details/8632430