Colored Sticks

/*Colored Sticks
You are given a bunch of wooden sticks. Each endpoint of each stick is colored with some color. Is it possible to align the sticks
in a straight line such that the colors of the endpoints that touch are of the same color?
Input
Input is a sequence of lines, each line contains two words, separated by spaces, giving the colors of the endpoints of one stick. A
word is a sequence of lowercase letters no longer than 10 characters. There is no more than 250000 sticks.
Output
If the sticks can be aligned in the desired way, output a single line saying Possible, otherwise output Impossible.
Sample Input
blue red
red violet
cyan blue
blue magenta
magenta cyan
Sample Output
Possible*/
//题意：有一些木棍，在木棍的两端涂颜色，给出一些颜色，问两根木棍的两端颜色能否都一样

#include<stdio.h>#include<iostream>#include<string.h>  using namespace std;     const int large=500000;  //25W条棒子，有50W个端点   class TrieTree_Node   //字典树结点  {      public:          bool flag;   //标记到字典树从根到当前结点所构成的字符串是否为一个(颜色)单词          int id;     //当前颜色(结点)的编号          TrieTree_Node* next[27];            TrieTree_Node()   //initial          {              flag=false;              id=0;              memset(next,0,sizeof(next));  //0 <-> NULL          }  }root;   //字典树根节点   int color=0;  //颜色编号指针，最终为颜色总个数    int degree[large+1]={0};   //第id个结点的总度数  int ancestor[large+1];   //第id个结点祖先   /*寻找x结点的最终祖先*/   int find(int x)  {      if(ancestor[x]!=x)          ancestor[x]=find(ancestor[x]);   //路径压缩      return ancestor[x];  }   /*合并a、b两个集合*/  void union_set(int a,int b)  {      int pa=find(a);      int pb=find(b);      ancestor[pb]=pa;   //使a的祖先 作为 b的祖先      return;  }    //利用字典树构造字符串s到编号int的映射   int hash(char *s)    {      TrieTree_Node* p=&root;  //从TrieTree的根节点出发搜索单词(单词不存在则创建)      int len=0;      while(s[len]!='\0')      {          int index=s[len++]-'a';  //把小写字母a~z映射到数字的1~26，        //作为字典树的每一层的索引           if(!p->next[index])  //当索引不存在时，构建索引              p->next[index]=new TrieTree_Node;          p=p->next[index];      }      if(p->flag)  //颜色单词已存在          return p->id;  //返回其编号      else   //否则创建单词      {          p->flag=true;          p->id=++color;          return p->id;   //返回分配给新颜色的编号      }  }  int main()    {      /*Initial the Merge-Set*/      for(int k=1;k<=large;k++)   //初始化，每个结点作为一个独立集合          ancestor[k]=k;  //对于只有一个结点x的集合，x的祖先就是它本身      /*Input*/      char a[11],b[11];      while(cin>>a>>b)        {          /*Creat the TrieTree*/            int i=hash(a);          int j=hash(b);  //得到a、b颜色的编号           /*Get all nodes' degree*/          degree[i]++;          degree[j]++;   //记录a、b颜色出现的次数(总度数)           /*Creat the Merge-Set*/           union_set(i,j);      }       /*Judge the Euler-Path*/       int s=find(1);  //若图为连通图，则s为所有结点的祖先                          //若图为非连通图，s为所有祖先中的其中一个祖先      int num=0;  //度数为奇数的结点个数       for(int i=1;i<=color;i++)      {          if(degree[i]%2==1)              num++;                          if(num>2)   //度数为奇数的结点数大于3，欧拉路必不存在          {              cout<<"Impossible"<<endl;              return 0;          }           if(find(i)!=s)   //存在多个祖先，图为森林，不连通          {              cout<<"Impossible"<<endl;              return 0;          }      }        if(num==1) //度数为奇数的结点数等于1，欧拉路必不存在          cout<<"Impossible"<<endl;      else       //度数为奇数的结点数恰好等于2或不存在，存在欧拉路          cout<<"Possible"<<endl;      return 0;  }