Codeforces 858F ( Codeforces Round #434 Div. 2 F ) Wizard's Tour 图论dfs

D. Wizard's Tour

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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 ai, bi(1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi 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.

Output

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.

Examples

input

4 51 23 22 43 44 1

output

21 4 24 3 2

input

5 85 31 24 55 12 54 31 43 2

output

41 4 52 3 41 5 35 2 1

要求在多个无向图当中选择最多的连接三个点的边对，每条边只能选一次。

有一种构造方法，可以使得每个连通块当中至多一条边不被选。构造方法如下：

先在图中随意dfs一棵生成树出来，再从叶子向上构造。

对于每个点，可以把与它相连的所有边分为两类：生成树上的边和其他边。现在，对于一个点，当与其相连未选的边数量为偶数时，我们就用这所有的n条边构造n/2个答案，中间点全部选现在的这个点即可；否则，我们把连向父亲的边甩给父亲，将剩下的n-1条边构造(n-1)/2个答案。这样，除了根节点上可能剩下一条边，其余的所有边都可以构造出答案。

不得不服出题人的思维。。。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <stack>#include <iomanip>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;typedef double db;const int maxn=200005, inf = 0x3f3f3f3f;const ll llinf = 0x3f3f3f3f3f3f3f3f;const ld pi = acos(-1.0L);vector<int> t[maxn];int ans[maxn][3],head[maxn];bool visit[maxn],use[maxn];int cnt,num;struct Edge {int from,to,pre;};Edge edge[maxn*2];void addedge(int from,int to) {edge[num]=(Edge){from,to,head[from]};head[from]=num++;edge[num]=(Edge){to,from,head[to]};head[to]=num++;}void dfs(int now,int fa,int pre) {visit[now] = 1;int size;for (int i=head[now];i!=-1;i=edge[i].pre) {int to=edge[i].to;if (to==0) continue;edge[i].to=edge[i^1].to=0;if (!visit[to]) {dfs(to, now, 1);if (!use[to]) t[now].push_back(to);} else t[now].push_back(to);}size = t[now].size() + pre;if (size==0) return;if (size % 2 == 0) {use[now] = 1;for (int i = 0; i+1 < size-pre; i+=2) ans[++cnt][0] = t[now][i], ans[cnt][1] = now, ans[cnt][2] = t[now][i+1];if (pre) ans[++cnt][0] = t[now][size-pre-1], ans[cnt][1] = now, ans[cnt][2] = fa;}else for (int i = 0; i + 1 < size-pre; i += 2)ans[++cnt][0] = t[now][i], ans[cnt][1] = now, ans[cnt][2] = t[now][i + 1];}int main() {int n, m, i, j, x, y;num=0;memset(head,-1,sizeof(head));scanf("%d%d", &n, &m);for (i = 1; i <= m; i++) {scanf("%d%d", &x, &y);addedge(x,y);}mem0(visit); cnt = 0; mem0(use);for (i = 1; i <= n; i++) if (!visit[i]) dfs(i, 0, 0);printf("%d\n", cnt);for (i = 1; i <= cnt; i++) printf("%d %d %d\n", ans[i][0], ans[i][1], ans[i][2]);return 0;}