UVa 108

题目来源：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=44

 Maximum Sum 

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

解题思路：

题意：给出n*n的矩阵，求出里面子矩阵的和的最大值。

最大连续子序列的应用，序列是一维的，矩阵是二维的，所以我们可以把矩阵转换为一维的来算。

也就是枚举矩阵的连续几行的合并，这样就转换为一维的了，再用最大子序列的算法去求，更新最大值就可以了。

代码：

#include <bits/stdc++.h>

using namespace std;

int table[100][100];
int sum[100];
int N;

int max_continuous_sum()
{
    int maxs=0,s=0;
    for(int i=0; i<N; i++)
    {
        if(s>=0) s+=sum[i];
        else s=sum[i];
        maxs = maxs>s ? maxs : s;
    }
    return maxs;
}
int main()
{
    cin >> N;
    int maxsum=0;
    int tmp;
    for(int i=0; i<N; i++)
    {
        for(int j=0; j<N; j++)
        {
            cin >> table[i][j];
            sum[j]=table[i][j];
        }
        tmp = max_continuous_sum();
        maxsum = maxsum>tmp ? maxsum : tmp;
        for(int j=i-1; j>=0; j--)
        {
            for(int k=0; k<N; k++)
                sum[k]+=table[j][k];
            tmp = max_continuous_sum();
            maxsum = maxsum>tmp ? maxsum : tmp;
        }
    }
    cout << maxsum << endl;
    return 0;
}