题目链接:

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=995

Problem D: The Necklace 

My little sister had a beautiful necklace made of colorful beads. Two successive beads in the necklace shared a common color at their meeting point. The figure below shows a segment of the necklace:

But, alas! One day, the necklace was torn and the beads were all scattered over the floor. My sister did her best to recollect all the beads from the floor, but she is not sure whether she was able to collect all of them. Now, she has come to me for help. She wants to know whether it is possible to make a necklace using all the beads she has in the same way her original necklace was made and if so in which order the bids must be put.

Please help me write a program to solve the problem.

Input

The input contains T test cases. The first line of the input contains the integer T.

The first line of each test case contains an integer N ( ) giving the number of beads my sister was able to collect. Each of the next N lines contains two integers describing the colors of a bead. Colors are represented by integers ranging from 1 to 50.

Output

For each test case in the input first output the test case number as shown in the sample output. Then if you apprehend that some beads may be lost just print the sentence ``some beads may be lost" on a line by itself. Otherwise, print N lines with a single bead description on each line. Each bead description consists of two integers giving the colors of its two ends. For , the second integer on line i must be the same as the first integer on line i + 1. Additionally, the second integer on line N must be equal to the first integer on line 1. Since there are many solutions, any one of them is acceptable.

Print a blank line between two successive test cases.

Sample Input

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

Sample Output

Case #1
some beads may be lost Case #2
2 1
1 3
3 4
4 2
2 2

这题就是判断是否存在欧拉回路。

每个点的度数必须为偶数,而且连通。

把颜色当成一个点。

递归打印路径。

//============================================================================
// Name : UVA.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================ #include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <map>
#include <vector>
using namespace std;
const int MAXN=;
int F[];
int find(int x)
{
if(F[x]==-)return x;
else return F[x]=find(F[x]);
}
void bing(int x,int y)
{
int t1=find(x);
int t2=find(y);
if(t1!=t2)F[t1]=t2;
}
int num[];
int G[][];
void Traverse(int u)
{
for(int v=;v<=;v++)
if(G[u][v]>)
{
G[u][v]--;
G[v][u]--;
Traverse(v);
printf("%d %d\n",v,u);
}
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int T;
int n;
scanf("%d",&T);
int iCase=;
while(T--)
{
if(iCase>)printf("\n");
iCase++;
scanf("%d",&n);
int u,v;
memset(F,-,sizeof(F));
memset(num,,sizeof(num));
memset(G,,sizeof(G));
for(int i=;i<n;i++)
{
scanf("%d%d",&u,&v);
num[u]++;
num[v]++;
bing(u,v);
G[u][v]++;
G[v][u]++;
}
bool flag=true;
int temp=-;
for(int i=;i<=;i++)
{
if(num[i]==)continue;
if(num[i]%)
{
flag=false;
break;
}
if(temp==-)
{
temp=find(i);
continue;
}
if(temp!=find(i))
{
flag=false;
break;
}
}
printf("Case #%d\n",iCase);
if(!flag)
{
printf("some beads may be lost\n");
continue;
}
for(int i=;i<=;i++)
if(num[i]!=)
{
u=i;
break;
}
Traverse(u);
}
return ;
}

UVA 10054 The Necklace(欧拉回路,打印路径)的更多相关文章

  1. UVA 10054 the necklace 欧拉回路

    有n个珠子,每颗珠子有左右两边两种颜色,颜色有1~50种,问你能不能把这些珠子按照相接的地方颜色相同串成一个环. 可以认为有50个点,用n条边它们相连,问你能不能找出包含所有边的欧拉回路 首先判断是否 ...

  2. Uva 10054 欧拉回路 打印路径

    看是否有欧拉回路 有的话打印路径 欧拉回路存在的条件: 如果是有向图的话 1.底图必须是连通图 2.最多有两个点的入度不等于出度 且一个点的入度=出度+1 一个点的入度=出度-1 如果是无向图的话 1 ...

  3. 【欧拉回路】UVA - 10054 The Necklace

    题目大意: 一个环被切割成了n个小块,每个小块有头尾两个关键字,表示颜色. 目标是判断给出的n个小块能否重构成环,能则输出一种可行解(按重构次序输出n个色块的头尾颜色).反之输出“some beads ...

  4. UVA 1626 区间dp、打印路径

    uva 紫书例题,这个区间dp最容易错的应该是(S)这种匹配情况,如果不是题目中给了提示我就忽略了,只想着左右分割忘记了这种特殊的例子. dp[i][j]=MIN{dp[i+1][j-1] | if( ...

  5. UVA 624 (0 1背包 + 打印路径)

    #include<stdio.h> #include<string.h> #include<stdlib.h> #include<ctype.h> #i ...

  6. UVA 531 - Compromise(dp + LCS打印路径)

      Compromise  In a few months the European Currency Union will become a reality. However, to join th ...

  7. uva 10054 The Necklace(欧拉回路)

    The Necklace  My little sister had a beautiful necklace made of colorful beads. Two successive beads ...

  8. UVa 10054 The Necklace(无向图欧拉回路)

    My little sister had a beautiful necklace made of colorful beads. Two successive beads in the neckla ...

  9. UVA 10054 The Necklace (无向图的欧拉回路)

    本文链接:http://www.cnblogs.com/Ash-ly/p/5405904.html 题意: 妹妹有一条项链,这条项链由许多珠子串在一起组成,珠子是彩色的,两个连续的珠子的交汇点颜色相同 ...

随机推荐

  1. Altium designer总结

    itwolf原创文章,转载请注明出处 大概有半年没有画过PCB板了,最近突然又要画一个简单的小板子,却发现好多东西已经不是很熟练了,现在把Altium designer软件的使用中要注意的问题和一些小 ...

  2. 【转载】React入门-Todolist制作学习

    我直接看的这个React TodoList的例子(非常好!): http://www.reqianduan.com/2297.html 文中示例的代码访问路径:http://127.0.0.1:708 ...

  3. fil_space_t

    typedef struct fil_space_struct fil_space_t; /** Tablespace or log data space: let us call them by a ...

  4. UVa 437 (变形的LIS) The Tower of Babylon

    题意: 有n种类型的长方体,每种长方体的个数都有无限个.当一个长方体的长和宽分别严格小于另一个长方体的长和宽的时候,才可以把这个放到第二个上面去.输出这n种长方体能组成的最大长度. 分析: 虽说每种都 ...

  5. BZOJ_1624_ [Usaco2008_Open]_Clear_And_Present_Danger_寻宝之路_(最短路_Floyd)

    描述 http://www.lydsy.com/JudgeOnline/problem.php?id=1025 给出\(n\)个点以及之间的边的长度,给出必须访问的点的顺序,求最短路线长度. 分析 用 ...

  6. 从客户端中检测到有潜在危险的 Request.Form 值-解决方案

    环境:VS2010 1.页头上加上ValidateRequest="false" <%@ Page Language="C#" ValidateReque ...

  7. 监听某个div或其它标签的大小改变来执行相应的处理

    jquery 默认的resize只能监听到浏览器窗口大小的改变,但我们在实际使用过程中有可能还需要监听某个div或其它标签的大小改变来执行相应的处理,如果使用默认的resize就无能为力了.怎么办呢, ...

  8. 《C++ Primer 4th》读书笔记 第8章-标准IO库

    原创文章,转载请注明出处:http://www.cnblogs.com/DayByDay/p/3936457.html

  9. System.arraycopy方法

    数组的复制有多种方法,其中有一种就是System.arraycopy方法,传闻速度也很快. 方法完整签名: public static void arraycopy(Object src, int s ...

  10. Struts2中通配符

    1.Struts2中通配符可通过请求的url路径来确定包.类.方法.返回值名. 如 <action name="*_*_*_*" class="cn.javass. ...