无向图求欧拉回路：

1、图连通

        2、所有顶点的度数位偶数

随便从一个点开始递归遍历即可求出路径

#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;
const int maxcolor = 50;
int n, G[maxcolor+1][maxcolor+1], deg[maxcolor+1];

struct Edge{
	int from, to;
	Edge(int from, int to):from(from), to(to){

	}
};

vector<Edge> ans;
void euler(int u){
	for(int v=1; v<=maxcolor; v++) if(G[u][v]){
		G[u][v]--; G[v][u]--;
		euler(v);
		ans.push_back(Edge(u, v));
	}
}

int main(){
	int T;
	scanf("%d", &T);
	for(int cas=1; cas<=T; ++cas){
		scanf("%d", &n);
		memset(G, 0, sizeof G );
		memset(deg, 0, sizeof deg );
		int start = -1;
		for(int i=0; i<n; ++i){
			int u, v;
			scanf("%d%d", &u, &v);
			G[u][v]++; G[v][u]++;
			deg[u]++; deg[v]++;
			start = u;
		}
		
		//无向图的欧拉回路
		bool solved = true;
		for(int i=1; i<=maxcolor; ++i)
			if(deg[i]%2==1) {
				solved = false; break;
			}
		if(solved){
			ans.clear();
			euler(start);
			if(ans.size() !=n || ans[0].to != ans[ans.size()-1].from)solved = false;
		}

		printf("Case #%d\n", cas);
		if(!solved){
			printf("some beads may be lost\n");
		}
		else for(int i=ans.size()-1; i>=0; i--) 
			printf("%d %d\n", ans[i].from, ans[i].to);
		if(cas<T) printf("\n");
	}
	return 0;
}

UVa10054 The Necklace，无向图求欧拉回路