并查集：Forming Teams(CF#133)

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在一场足球比赛中，n名学生需要分成两队，但存在互为敌人的关系。每个学生最多有两个敌人，敌人关系是双向的。目标是确保没有两个敌人在同一队，若无法避免，则需部分学生坐替补席。输入包括学生数n和敌人对数m，以及具体的敌人关系。题目要求找到最少需要多少学生坐替补席以满足分组条件。输出为一个整数，表示最少的替补人数。

Forming Teams

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — 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

Output

Input

6 2
1 4
3 4

Output

Input

Output

2

出现奇数环的时候踢一个人

#include <iostream>
#include <cstdio>

using namespace std;

int sum[110], pa[110];
int n;

void init(){
    for(int i = 1; i <= n; i ++){
        sum[i] = 1;
        pa[i] = i;
    }
}

int find(int i){
    if(pa[i] == i)
        return i;
    else
        return find(pa[i]);
}

int main(){
    int m;
    int i, j, k;
    int ans;
    cin >> n >> m;
    ans = n;
    init();
    for(i = 0; i < m; i ++){
        int a, b;
        cin >> a >> b;
        a = find(a);
        b = find(b);
        if(a != b){
            pa[a] = b;
            sum[b] += sum[a];
        }
        else{
            if(sum[a] % 2)
                ans --;
        }
    }
    if(ans % 2)
    ans --;
    cout << n - ans << endl;
    return 0;
}