## Tuesday, September 30, 2014

### UVa Problem 10505 - Montesco vs Capuleto

Problem:

Solution:

The problem indicated the graph defined by the is enemy of relationship is a bipartite graph, and the problem is about finding a maximum independent set in this graph. That is the same as finding the connected component and choosing the maximum side. By 2-coloring the graph we can easy complete the problem.

The 'real' problem with this problem is that the test cases are bad. The test cases does not really follow the rules. The graph could be non-bipartite and the input have edges linking to nowhere (i.e. outside of the node set). One need to carefully dealt with that two bad inputs by correctly ignoring them.

What I learnt:

• Problem input and be bad - use assert to make sure if in doubt.

Code:

#include "stdafx.h"

// http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1361

#include "UVa10505.h"

#include <iostream>
#include <list>
#include <map>
#include <queue>

using namespace std;

int UVa10505()
{
int number_of_cases;
cin >> number_of_cases;
for (int c = 0; c < number_of_cases; c++)
{
list<pair<int, int> > input;

int number_of_people;
cin >> number_of_people;
for (int i = 1; i <= number_of_people; i++)
{
int number_of_enemies;
cin >> number_of_enemies;
for (int e = 0; e < number_of_enemies; e++)
{
int enemy;
cin >> enemy;
if (1 <= enemy && enemy <= number_of_people) // This should never happen - but who knows - the bipartite condition is broken - this might as well happen.
{
input.push_back(pair<int, int>(i, enemy));
}
}
}

for (list<pair<int, int> >::iterator ii = input.begin(); ii != input.end(); ii++)
{
int from = ii->first;
int to = ii->second;
from--;
to--;
}

int* visited = new int[number_of_people];
for (int i = 0; i < number_of_people; i++)
{
visited[i] = 0;
}
for (int i = 0; i < number_of_people; i++)
{
int trueCount = 0;
int falseCount = 0;
if (visited[i] == 0)
{
queue<pair<int, int> > bfsQueue;
bfsQueue.push(pair<int, int>(i, 1));
visited[i] = 1;
while (bfsQueue.size() > 0)
{
pair<int, int> visiting = bfsQueue.front();
bfsQueue.pop();
if (visiting.second == 1)
{
trueCount++;
}
else
{
falseCount++;
}
{
if (visited[*ni] == 0)
{
visited[*ni] = -visiting.second;
bfsQueue.push(pair<int, int>(*ni, -visiting.second));
}
else if (visited[*ni] == visiting.second)
{
}
}
}
{
}
}
}

}