## Monday, November 3, 2014

### UVa Problem 10911 - Forming Quiz Teams

Problem:

Solution:

This problem is generally known as the minimum cost perfect matching problem. It is a hard (but solved) problem. For this problem, however, the number of nodes is small, we can use dynamic programming instead. The key insight is that after a pair is selected, the rest must also be optimally paired up. So we have the optimal sub-structure. If six students are selected, no matter how they are paired, the rest must also be optimally paired up independent on how the six are paired, so the sub-problems are overlapped.

By introducing a memo, we can use top-down DP to solve the problem. Implementation wise, we use bitmap to compactly encode the name of a sub-problem. With that, please see the code below.

Code:

#include "stdafx.h"

// http://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=1852

#include "UVa10911.h"

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <iomanip>

using namespace std;

double find_minimum_cost(int num_students, int problem, double** distance, double* memo)
{
double result = memo[problem];
if (result != -1)
{
return result;
}
else
{
vector<int> unpaired_students;
for (int i = 0; i < num_students; i++)
{
int student_bit_mask = 1 << i;
if ((problem & student_bit_mask) != 0)
{
unpaired_students.push_back(i);
}
}

bool first = true;
double best = -1;

for (unsigned int i = 0; i < unpaired_students.size(); i++)
{
int first_bit_mask_on = 1 << unpaired_students[i];
for (unsigned int j = i + 1; j < unpaired_students.size(); j++)
{
int second_bit_mask_on = 1 << unpaired_students[j];
double candidate = find_minimum_cost(num_students, problem, distance, memo) + distance[unpaired_students[i]][unpaired_students[j]];
if (first)
{
best = candidate;
first = false;
}
else
{
if (candidate < best)
{
best = candidate;
}
}
}
}

memo[problem] = best;
return best;
}
}

int UVa10911()
{
int case_number = 1;
while (true)
{
int num_pairs;
cin >> num_pairs;
if (num_pairs == 0)
{
break;
}
int num_students = num_pairs * 2;
vector<pair<int, int> > student_locations;

for (int i = 0; i < num_students; i++)
{
string name;
int x;
int y;
cin >> name;
cin >> x;
cin >> y;
student_locations.push_back(pair<int, int>(x, y));
}

// Step 1: Compute distance matrix to avoid re-computation
double** distance = new double*[num_students];
for (int i = 0; i < num_students; i++)
{
distance[i] = new double[num_students];
}

for (int i = 0; i < num_students; i++)
{
distance[i][i] = 0;
for (int j = i + 1; j < num_students; j++)
{
double x_diff = (student_locations[i].first - student_locations[j].first);
double y_diff = (student_locations[i].second - student_locations[j].second);
distance[i][j] = distance[j][i] = sqrt(x_diff * x_diff + y_diff * y_diff);
}
}

int table_size = 1 << num_students;

double* memo = new double[table_size];
for (int i = 0; i < table_size; i++)
{
// Representing the sub-problem is not solved
memo[i] = -1;
}
// The sub-problem of selecting no pair is trivially solved - the cost is 0
memo[0] = 0;

std::cout << std::setprecision(2) << std::fixed;
cout << "Case " << case_number << ": " << find_minimum_cost(num_students, table_size - 1, distance, memo) << endl;

for (int i = 0; i < num_students; i++)
{
delete[] distance[i];
}
delete[] distance;
case_number++;
}

return 0;
}