## Monday, November 3, 2014

### UVa Problem 103 - Stacking Boxes

Problem:

Solution:

First of all - it is easy to determine if a box nest in another - all we need to do is to sort the dimensions and make sure it fits in our dimensions.

With that, with an $O(n^2)$ loop, it is easy to build a directed acyclic graph (DAG) of boxes when one node represent a box and an arrow represent the "can contain" relationship.

Then we are asked to find the longest sequence. So it is simply to longest path in the DAG. To practice dynamic programming, I used the longest increasing subsequence method. First, I topologically sorted the graph, then I find a longest increasing subsequence in the topological order, that gives the sequence.

Code:

#include "stdafx.h"

// http://uva.onlinejudge.org/index.php?option=onlinejudge&Itemid=8&page=show_problem&problem=39

#include "UVa103.h"

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

int UVa103()
{
while (true)
{
int number_of_boxes;
int dimension;
cin >> number_of_boxes;
if (cin.eof())
{
break;
}
cin >> dimension;
vector<vector<int> > boxes;
boxes.resize(number_of_boxes);
for (int i = 0; i < number_of_boxes; i++)
{
boxes[i].resize(dimension);
for (int d = 0; d < dimension; d++)
{
int length;
cin >> length;
boxes[i][d] = length;
}
sort(boxes[i].begin(), boxes[i].end());
}

// Step 1: Build DAG - larger box point to smaller box (so that the reverse longest path is smaller nested within largest)
vector<int> indegree;
indegree.resize(number_of_boxes);
for (int i = 0; i < number_of_boxes; i++)
{
indegree[i] = 0;
}
for (int i = 0; i < number_of_boxes; i++)
{
for (int j = i + 1; j < number_of_boxes; j++)
{
int global_compare = 2;
for (int d = 0; d < dimension; d++)
{
int local_compare;
if (boxes[i][d] == boxes[j][d])
{
local_compare = 0;
}
else if (boxes[i][d] > boxes[j][d])
{
local_compare = 1;
}
else
{
local_compare = -1;
}
if (global_compare == 2)
{
global_compare = local_compare;
}
else if (global_compare != local_compare)
{
global_compare = 0;
break;
}
}
if (global_compare == 1)
{
indegree[j]++;
}
else if (global_compare == -1)
{
indegree[i]++;
}
}
}

// Step 2: Topological sort
queue<int> nodes_with_zero_degree;
for (int i = 0; i < number_of_boxes; i++)
{
if (indegree[i] == 0)
{
nodes_with_zero_degree.push(i);
}
}
vector<int> topological_order;
while (nodes_with_zero_degree.size() > 0)
{
int current_node = nodes_with_zero_degree.front();
nodes_with_zero_degree.pop();
topological_order.push_back(current_node);
{
int neighbor = *ni;
if (--indegree[neighbor] == 0)
{
nodes_with_zero_degree.push(neighbor);
}
}
}

// Step 3: Find longest path in this sequence
vector<int> all_longest_path_length;
vector<int> all_longest_path_parent;
all_longest_path_length.resize(number_of_boxes);
all_longest_path_parent.resize(number_of_boxes);
all_longest_path_length[0] = 1;
all_longest_path_parent[0] = -1;

int longest_path_length = -1;
int longest_path_index = -1;

for (int i = 1; i < number_of_boxes; i++)
{
int best_path_length = 1;
int best_path_index = -1;
for (int j = 0; j < i; j++)
{
int last_path_length;
last_path_length = all_longest_path_length[j];
{
int new_path_length = last_path_length + 1;

if (new_path_length > best_path_length)
{
best_path_length = new_path_length;
best_path_index = j;
}
}
}

if (best_path_length > longest_path_length)
{
longest_path_length = best_path_length;
longest_path_index = i;
}

all_longest_path_length[i] = best_path_length;
all_longest_path_parent[i] = best_path_index;
}

cout << longest_path_length << endl;
int longest_path_cursor = longest_path_index;
while (longest_path_cursor != -1)
{
cout << (topological_order[longest_path_cursor] + 1);
longest_path_cursor = all_longest_path_parent[longest_path_cursor];
if (longest_path_cursor != -1)
{
cout << " ";
}
}

cout << endl;

}
return 0;
}