uva 10054 项链

原题：

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 (5 ≤ N ≤ 1000) 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 1 ≤ i ≤ N1, 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

题目大意：

给一堆珠子，每个珠子分为两头各有一个一种颜色。现在让你把这些珠子串成一个圈，使得项链的两个珠子的相邻两端颜色相同。然后把顺序打印出来~

注意，每个珠子可能有种~

#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<vector>
#include<cmath>
#include<stack>
#include<queue>
#include<iomanip>
using namespace std;
int gp[51][51];
int mark[51];
void euler(int head)
{
for(int i=0;i<=50;i++)
{
if(gp[head][i])
{
gp[head][i]--;
gp[i][head]--;;
euler(i);
cout<<i<<" "<<head<<endl;//这里注意打印方向不要反
}
}
}
void ini()
{
memset(gp,0,sizeof(gp));
memset(mark,0,sizeof(mark));
}
int main()
{
ios::sync_with_stdio(false);
int a,b,n,k,t=0,e;
cin>>k;
while(k--)
{
e=1;
cin>>n;
ini();
for(int i=0;i<n;i++)
{
cin>>a>>b;
gp[a]++;
gp[b][a]++;;
mark[a]++;
mark[b]++;
}
for(int i=0;i<=50;i++)
if(mark[i]%2)//判断能否成为回一个欧拉回路
{
e=0;break;
}
if(t)
cout<<endl;
if(e)
{
cout<<"Case #"<<++t<<endl;
euler(a);
}
else
{
cout<<"Case #"<<++t<<endl;
cout<<"some beads may be lost"<<endl;
}
}
return 0;
}

[b]思路：

拿着刘汝佳的白书随便挑了一个就开始做，挑中这道题。很明显是一个环形一笔画的问题。

刚开始做的时候我把打印的路径都存在一个deque里面，结果居然超时。只能直接打印路径，这里面刘汝佳白书的那个模板不错，注意打印路径是一个环，所以进行euler判断的时候不用先把第一个元素标记过，否则没有办法打印出开头那个珠子。

此外，这里相同的珠子可能有很多个，所以不能光用vis判断，应该在把每个珠子计数，使用过一个以后就gp[a][b]–即可。