Topcoder SRM 709 DIV 2 500pt Permatchd2 solution

Problem Statement
	Hero has a simple undirected graph. (Simple means that each edge connects two different vertices, and each pair of vertices is connected by at most one edge.) A graph is considered pretty if it is a simple undirected graph in which each connected component contains an even number of edges. You are given the adjacency matrix of Hero's graph as a vector <string> graph. ('Y' means that the two vertices are connected by an edge, 'N' means that they aren't.) Change Hero's graph into a pretty graph by adding as few edges as possible. Return the minimum number of edges you have to add, or -1 if Hero's graph cannot be changed into a pretty graph.
Definition
	Class: Permatchd2 Method: fix Parameters: vector <string> Returns: int Method signature: int fix(vector <string> graph) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 256 Stack limit (MB): 256
Notes
-	Don't forget that the graph you'll obtain after adding the edges must still be simple.
Constraints
-	n will be between 1 and 50, inclusive.
-	graph will contain exactly n elements.
-	Each element in graph will contain exactly n characters.
-	Each character in graph will be either 'N' or 'Y'.
-	For all valid i graph[i][i] will be equal to 'N'.
-	For all valid i, j graph[i][j] will be equal to graph[j][i].
Examples
0)
	{"NYN", "YNN", "NNN"} Returns: 1 This is the adjacency matrix of a simple graph on three vertices. Two of the three vertices are connected by an edge. Here, an optimal solution is to add one edge that will connect the remaining vertex to one of the other two. The resulting graph is a simple graph with a single connected component. The connected component contains two edges, which is an even number.
1)
	{"NYY", "YNN", "YNN"} Returns: 0
2)
	{"NYY", "YNY", "YYN"} Returns: -1
3)
	{"NYYY", "YNYY", "YYNN", "YYNN"} Returns: 1
4)
	{"NYNNNN", "YNNNNN", "NNNYNN", "NNYNNN", "NNNNNY", "NNNNYN"} Returns: 3
5)
	{"N"} Returns: 0

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Solution:

  The size of the number of nodes is less than 50, which means we can use direct search on graph.

  First, we build the graph.

  Second, search all the nodes that have not been searched. Create unions with edge and nodes numbers.

  Then we search all the unions and record the number of ugly unions(unions that have odd number of edges).

   If there's only one ugly union in the whole graph, we want to see if all the nodes are connected. If not, we could connect any pair of unconnected nodes to make it a pretty graph. print "1".

   If there exists at least one union with even number of edges, we could create a pretty graph by connecting all ugly unions to this pretty union. print the number of ugly unions.

   If there's only ugly unions in the graph, we connect all of them together and add one edge to any pair of unconnected nodes to make it a pretty graph. print the number of ugly unions - 1 + 1 = number of ugly unions.

  The interesting magic of this problem is that we can reduce the cases to two: only one ugly union & exists at least one pretty union.

Codes:

// BEGIN CUT HERE

// END CUT HERE
#line 5 "Permatchd2.cpp"
#include<bits/stdc++.h>
using namespace std;
int a[55][55];
int size[55];
int edge, node;
int vis[55];
int kk[55], gg[55];
int ss;
bool flag;
void dfs(int root){
 vis[root] = 1;
 edge += size[root];
 for(int i = 1; i <= size[root]; i++){
  if(!vis[a[root][i]]){
   node++;
   dfs(a[root][i]);
  }
 }
}
class Permatchd2 {
 public:
 int fix(vector <string> graph) {
  int n = graph.size();
  for(int i = 0; i < n; i++){
   for(int j = 0; j < n; j++){
    if(graph[i][j] == 'Y') size[i]++, a[i][size[i]] = j;
   }
  }
  for(int i = 0; i < n; i++){
   edge = 0, node = 1;
   if(!vis[i]){
    ss++;
    dfs(i);
    kk[ss] = edge / 2;
    gg[ss] = node;
    if(kk[ss] < node * (node - 1) / 2){
     flag = 1;
    }
   }
  }
  int count = 0;
  for(int i = 1; i <= ss; i++){
   if(kk[i] % 2 == 1) count++;
  }
  //return count;
  int ans = 0, remain = ss - count;
  if(ss == 1 && count == 1) {
   if(flag) return 1;
   else return (-1);
  }
  else return count;
 }
};