Codeforces Round #133 (Div. 2) B. Forming Teams（并查集）

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本文介绍了一个算法问题，即如何将有互斥关系的学生公平地分为两支人数相等的足球队伍，若无法实现，则需计算最少需让几位学生离场观战。该问题通过并查集的数据结构解决，确保没有互为仇敌的学生被分配到同一队。

One day n students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.

We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student A is an archenemy to student B, then student B is an archenemy to student A.

The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.

Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last.

Input

The first line contains two integers n and m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of students and the number of pairs of archenemies correspondingly.

Next m lines describe enmity between students. Each enmity is described as two numbers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.

You can consider the students indexed in some manner with distinct integers from 1 to n.

Output

Print a single integer — the minimum number of students you will have to send to the bench in order to start the game.

Sample test(s)

input
5 4
1 2
2 4
5 3
1 4

output
1

input
6 2
1 4
3 4

output
0

input
6 6
1 2
2 3
3 1
4 5
5 6
6 4

output
2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m,i,j,k,p[105],q[105],ans;
int find(int x)
 {
     if(p[x]==x)
       return x;
     else
     {
      p[x]=find(p[x]);
     return find(p[x]);
     }
 }

int main()
{
   int a,b,x,y;

 while(~scanf("%d%d",&n,&m))
   {
       ans=0;
     memset(p,0,sizeof(p));
     memset(q,0,sizeof(q));
     for(i=0;i<n;i++)
     {
         p[i]=i;q[i]=1;
     }
     for(i=0;i<m;i++)
      {
          scanf("%d%d",&a,&b);
            x=find(a);
            y=find(b);
          if(x==y)
           {
               if(q[x]%2==1)
                 ans++;
           }
          else
           {
               p[x]=y;
               q[y]+=q[x];
           }
      }
     if((n-ans)%2==1)
       ans++;
     cout<<ans<<endl;

   }

  return 0;
}