【-30】858F. Wizard's Tour[dfs]

题目描述

All Berland residents are waiting for an unprecedented tour of wizard in his Blue Helicopter over the cities of Berland!

It is well-known that there are $n$ cities in Berland, some pairs of which are connected by bidirectional roads. Each pair of cities is connected by no more than one road. It is not guaranteed that the road network is connected, i.e. it is possible that you can’t reach some city from some other.

The tour will contain several episodes. In each of the episodes:

  • the wizard will disembark at some city $x$ from the Helicopter;
  • he will give a performance and show a movie for free at the city $x$;
  • he will drive to some neighboring city $y$ using a road;
  • he will give a performance and show a movie for free at the city $y$;
  • he will drive to some neighboring to $y$ city $z$;
  • he will give a performance and show a movie for free at the city $z$;
  • he will embark the Helicopter and fly away from the city $z$.

It is known that the wizard doesn’t like to use roads, so he agrees to use each road at most once (regardless of direction). In other words, for road between $a$ and $b$ he only can drive once from $a$ to $b$, or drive once from $b$ to $a$, or do not use this road at all.

The wizards wants to plan as many episodes as possible without violation the above rules. Help the wizard!

Please note that the wizard can visit the same city multiple times, the restriction is on roads only.

输入输出格式

输入格式

The first line contains two integers $n$, $m$ (1 ≤ $n$ ≤ 2·105, 0 ≤ $m$ ≤ 2·105) — the number of cities and the number of roads in Berland, respectively.

The roads description follow, one in each line. Each description is a pair of two integers $a$$i$, $b$$i$ (1 ≤ $a$$i$, $b$$i$ ≤ $n$, $a$$i$ ≠ $b$$i$), where $a$$i$ and $b$$i$ are the ids of the cities connected by the $i$-th road. It is guaranteed that there are no two roads connecting the same pair of cities. Every road is bidirectional. The cities are numbered from 1 to $n$.

It is possible that the road network in Berland is not connected.

输出格式

In the first line print $w$ — the maximum possible number of episodes. The next $w$ lines should contain the episodes in format $x$, $y$, $z$ — the three integers denoting the ids of the cities in the order of the wizard’s visits.

输入输出样例

输入样例1

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

输出样例1

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3
2
1 4 2
4 3 2

输入样例2

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2
3
4
5
6
7
8
9
5 8
5 3
1 2
4 5
5 1
2 5
4 3
1 4
3 2

输出样例2

1
2
3
4
5
4
1 4 5
2 3 4
1 5 3
5 2 1

题解

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#include<bits/stdc++.h>
using namespace std;
int read()
{
char s;
int k=0,base=1;
while((s=getchar())!='-'&&s!=EOF&&!(s>='0'&&s<='9'));
if(s==EOF)exit(0);
if(s=='-')base=-1,s=getchar();
while(s>='0'&&s<='9')
{
k=k*10+(s-'0');
s=getchar();
}
return k*base;
}
void write(int x)
{
if(x<0)
{
putchar('-');
write(-x);
}
else
{
if(x/10)write(x/10);
putchar(x%10+'0');
}
}
struct node
{
int lb;
int id[400100];
int to[400100];
int ne[400100];
int po[200100];
void add(int x,int y,int z)
{
lb++;
id[lb]=z;
to[lb]=y;
ne[lb]=po[x];
po[x]=lb;
}
} a;
int n,m,lb,X,Y;
int fa[200100];
bool vis[200100];
bool used[200100];
int ans[4][200100];
int deep[200100];
int Ans;
void dfs(int x,int last)
{
deep[x]=deep[fa[x]]+1;//记深度
vis[x]=true;
int num=0;
for (int i=a.po[x];i;i=a.ne[i])//先把所有连的没有遍历的点遍历一变
if (!vis[a.to[i]])
{
fa[a.to[i]]=x;//标记在搜索树上的父亲
dfs(a.to[i],i);
}
bool flag=1;
for (int i=a.po[x];i;i=a.ne[i])
{
if (used[a.id[i]]||deep[a.to[i]]<deep[x]) continue;//深度更小或者这条边用过了就退出
if (flag)//第一条边(旅行路线上的)
{
Ans++;
ans[1][Ans]=a.to[i];
ans[2][Ans]=x;
} else//第二条边(旅行路线上的)
{
ans[3][Ans]=a.to[i];
}
flag^=1;
used[a.id[i]]=1;//标记
}
if (!flag)//最后还有一条路线只有一条边,就把该点和父节点连的边加上去
{
ans[3][Ans]=fa[x];
if (fa[x]==0) Ans--;
used[a.id[last]]=1;//标记
}

}
int main()
{
freopen("graph.in","r",stdin);
freopen("graph.out","w",stdout);
n=read();m=read();
for (int i=1;i<=m;i++)
{
X=read();Y=read();
a.add(X,Y,i);
a.add(Y,X,i);
}
for (int i=1;i<=n;i++)
if (!vis[i]) dfs(i,0);//扫每个联通块长度
cout<<Ans<<'\n';
for (int i=1;i<=Ans;i++)
cout<<ans[1][i]<<" "<<ans[2][i]<<" "<<ans[3][i]<<endl;
return 0;
}