POJ 2230 Watchcow Euler circuit

Test instructions

For a diagram, a scheme that starts from point 1 and passes through all sides exactly two times to point 1, the data is guaranteed to have a solution.

Analysis:

Euler loop, change the number of times just fine.

Code:

POJ 2230
//sep9
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=10024;
const int maxm=50048;
int HEAD[MAXN];
int used[maxm*2];
vector<int> path;
struct Edge
{
	int v,nxt;	
} EDGE[MAXM*2];
int e,n,m;

void Dfs (int u)
{for
	(int i=head[u];i!=-1;i=edge[i].nxt) {
		int cur=i|1;
		if (used[cur]<2) {
			int v=edge[i].v;
			++used[cur];
			DFS (v);
			Path.push_back (v);

}} int main ()
{
	scanf ("%d%d", &n,&m);
	int i;
	e=0;
	memset (head,-1,sizeof (head));
	memset (used,0,sizeof (used));
	Path.clear ();
	for (i=1;i<=m;++i) {
		int u,v;
		scanf ("%d%d", &u,&v);
		edge[e].v=v;edge[e].nxt=head[u];head[u]=e++;
		edge[e].v=u;edge[e].nxt=head[v];head[v]=e++;
	}
	DFS (1);
	Reverse (Path.begin (), Path.end ());
	printf ("1\n");
	For (I=0;i<path.size (); ++i)
		printf ("%d\n", Path[i]);
	return 0;