## codeforces CF209C Trails and Glades 欧拉回路 转

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<center>time limit per test: 4 seconds</center> <center>memory limit per test: 256 megabytes</center> <center>input: standard input</center> <center>output: standard output</center>

Vasya went for a walk in the park. The park has $n$ glades, numbered from $1$ to $n$ . There are $m$ trails between the glades. The trails are numbered from $1$ to $m$ , where the $i$ -th trail connects glades $x_i$ and $y_i$. The numbers of the connected glades may be the same $(x_i=y_i)$, which means that a trail connects a glade to itself. Also, two glades may have several non-intersecting trails between them. 　 Vasya is on glade $1$ ,he wants to walk on all trails of the park exactly once, so that he can eventually return to glade $1$ . Unfortunately, Vasya does not know whether this walk is possible or not. Help Vasya, determine whether the walk is possible or not. If such walk is impossible, find the minimum number of trails the authorities need to add to the park in order to make the described walk possible. 　 Vasya can shift from one trail to another one only on glades. He can move on the trails in both directions. If Vasya started going on the trail that connects glades $a$ and $a$ , from glade $a$ , then he must finish this trail on glade $b$ .

### Input

The first line contains two integers $n$ and $m (1 \le n \le 10^6;0 \le m \le 10^6)$ — the number of glades in the park and the number of trails in the park, respectively. Next m lines specify the trails. The $i$-th line specifies the $i$-th trail as two space-separated numbers, $x_i,y_i (1\le x_i,y_i \le n)$ — the numbers of the glades connected by this trail.

### Output

Print the single integer — the answer to the problem. If Vasya's walk is possible without adding extra trails, print $0$, otherwise print the minimum number of trails the authorities need to add to the park in order to make Vasya's walk possible.

### Examples

#### Input1

 3 3
1 2
2 3
3 1


#### output1

 0


#### input2

 2 5
1 1
1 2
1 2
2 2
1 2


#### output2

 1


### Note

In the first test case the described walk is possible without building extra trails. For example, let's first go on the first trail, then on the second one, and finally on the third one. 　 In the second test case the described walk is impossible without adding extra trails. To make the walk possible, it is enough to add one trail, for example, between glades number one and two.

## 题目大意

• 最少添加多少条边，是无向图有欧拉回路。

• $n,m \le 10^6$

## 题解

• 求出每个点的度数

• 奇数点需要连一条新边

• 仅有偶度数点的连通块需要连两条新边

• 特判没有奇数点只有偶数点且只有一个连通块（已经成为欧拉回路）的情况，新边数为 $0$

• 答案为上面统计的新边数 $/ 2$

• 坑点：

• $1.$ 没有自环的孤立点不用连边，没有边相连的点可以不用到达

• $2.$ 但是$1$ 必须到达，所以特定要访问 $1$

• $3.$ 注意，自环不能直接删掉，孤立点的自环也要访问

## 代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int n,m,ans;
int deg[1000005],cnt;
vector<int>e[1000005];
bool vis[1000005],flag;
void dfs(int u){
vis[u]=1;
if(deg[u]&1) flag=1;
for(int i=0;i<e[u].size();++i)
if(!vis[e[u][i]]) dfs(e[u][i]);
}
bool pot[1000005],p1;
int main(){
scanf("%d %d",&n,&m);
for(int u,v,i=1;i<=m;++i){
scanf("%d %d",&u,&v);
pot[u]=pot[v]=1;
if(u==v) continue;
++deg[u]; ++deg[v];
e[u].push_back(v);
e[v].push_back(u);
}
pot[1]=1;
for(int i=1;i<=n;++i){
if(deg[i]&1){
++ans;
p1=1;
} else if(!vis[i]&&pot[i]){
++cnt;
flag=0;
dfs(i);
if(!flag) ans+=2;
}
}
if(!p1&&cnt==1) ans-=2;
printf("%d",ans/2);
return 0;
}

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