UVa 10054 - The Necklace (欧拉回路)

题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=995&mosmsg=Submission+received+with+ID+26578658

如果把每个珠子看成点，不好建图。

将每种颜色看成点，每个珠子就对应一条无向边，于是转化成求欧拉回路的问题。

无向图求欧拉回路要注意经过一条边以后，将反向边也标记。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 5010;
const int maxcolor = 50;

int T, n;

int h[maxcolor+10], cnt = 1;
struct e{
	int from, to, id, next;
}e[maxn]; 
void add(int u, int v, int id){
	e[++cnt].to = v;
	e[cnt].from = u;
	e[cnt].id = id; // 项链的编号
	e[cnt].next = h[u];
	h[u] = cnt; 
}

int vis[maxn], deg[maxcolor+10];
vector<int> path;

void euler(int u){
	for(int i = h[u] ; i != -1 ; i = e[i].next){
		if(!vis[i]){
			vis[i] = 1;
			vis[i ^ 1] = 1; 
			int v = e[i].to;
			euler(v);
			path.push_back(i);
		}
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	int kase = 0;
	int flag = 0; 
	T = read();
	while(T--){
		if(flag) printf("\n");
		flag = 1;
		path.clear();
		memset(deg, 0, sizeof(deg));
		memset(h, -1, sizeof(h)); cnt = 1;
		memset(vis, 0, sizeof(vis));
		
		n = read();
		
		int u, v;
		int start = -1;
		for(int i = 1 ; i <= n ; ++i){
			u = read(), v = read();
			++deg[u]; ++deg[v];
			add(u, v, i); add(v, u, i);
			start = u;
		}
		bool solved = true;
		
		for(int i = 1 ; i <= maxcolor ; ++i){
			if(deg[i] % 2 == 1){
				solved = false;
				break;
			}
		} 
		
		if(solved) euler(start);
		
		printf("Case #%d\n", ++kase);
		if(path.size() != n ||e[path[0]].to != e[path[path.size()-1]].from){
			solved = false;
		} 
		if(!solved) printf("some beads may be lost\n");
		else {
			for(int i = path.size() - 1 ; i >= 0 ; --i){
				printf("%d %d\n", e[path[i]].from, e[path[i]].to);
			}			
		}
	}
	return 0;
}