題意：判斷能否構成歐拉回路，不能的話輸出最小需要新增多少條邊

題解：並查集處理出聯通分量，輸入時處理出奇度數點個數，結果即為  奇度數點個數/2+不含奇度數點的聯通分量個數，1節點自動算作一個聯通分量 不管是否有邊連線

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <list>
#include <queue>
#include <map>
#include <stack>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-3
#define maxn 100010
#define MOD 1000000007

int n,m,ans;
int num[1000010],flag[1000010];
int fa[1000010],vis[1000010];
int Find(int x)
{
    return fa[x] = x == fa[x]?x:Find(fa[x]);
}
void Union(int x,int y)
{
    int fx = Find(x);
    int fy = Find(y);
    if(fx != fy)
        fa[fy] = fx;
}
int main()
{
    int t,C = 1;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        memset(num,0,sizeof(num));
        memset(flag,0,sizeof(flag));
        memset(vis,0,sizeof(vis));
        for(int i = 1; i <= n; i++)
            fa[i] = i;
        int u,v;
        vis[1] = 1;
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d",&u,&v);
            if(u != v)
            {
                num[u]++;
                num[v]++;
                vis[u]++;
                vis[v]++;
                Union(u,v);
            }
            else
                vis[u]++;
        }
        int ans = 0,k = 0,tong = 0;
        for(int i = 1; i <= n; i++)
            if(num[i]&1)
                flag[Find(i)]++;
        for(int i = 1; i <= n; i++)
            if(vis[i] && Find(i)== i)
            {
                if(flag[i])
                    ans += flag[i];
                else
                    tong++;
                k++;
            }
        if(k == 1)
            printf("%d\n",ans/2);
        else
            printf("%d\n",ans/2+tong);
    }
    return 0;
}