题目链接：Colored Sticks

一道3个知识点结合的题目，可以说单个知识点的题目，都会做，一旦知识点结合起来，题目就不简单了

思路：这个题开始看就知道是并查集，但是不好处理的不同种单词的统计，所以理所应当联想到字典树，上次做字典树的题目啸爷出的是统计相同单词数，这个题目和那个一样，把flag加个编号即可，再利用并查集。

1750ms  水过

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
const int N = 500001;
using namespace std;

int father[N];
int du[N];
char a[20],b[20];
struct node{
    int bianhao;
    node *next[26];
};
node *root;
int hao;

int findx(int r)
{
        while(father[r]!=r)
       {
           r = father[r];
       }
       int i = r,j;
       while(father[i]!=r)
       {
           j = father[i];
           father[i] = r;
           i = j;
       }
       return r;
}

void merge(int a,int b)
{
       int fx = findx(a);
       int fy = findx(b);

       if(fx != fy)
        father[fx]=fy;
}
struct node *make()
{
       node *p;
       p = (struct node*)malloc(sizeof(struct node));
       for(int i = 0;i < 26;i++)
       {
            p->next[i] = NULL;
       }

       p->bianhao = 0;
       return p;
}
int INsert(char *s)
{
       node *p = root;
       int n = 0;
       int len = strlen(s);
       for(int i = 0;i<len;i++)
       {
           n = s[i]-'a';
          if(p->next[n] == NULL)
          {
               p->next[n] = make();
          }

          p = p->next[n];
       }
       if(p->bianhao == 0)
       {
           hao++;
           p->bianhao = hao;
       }

       return p->bianhao;
}
int ans;
bool flag;
void oula()
{
      for(int i = 1;i<=hao;i++)
       {
         if(du[i]%2==1)
            ans++;
       }
       for(int j=2;j<=hao;j++)
       {
         if(findx(j)!=findx(1))
            {
                flag = 1;
               break;
            }
       }
}
void make_root()
{
      root=(struct node*)malloc(sizeof(struct node));
       for(int i = 0;i<26;i++)
       {
           root->next[i] = NULL;
       }
       root->bianhao = 0;
}
int main()
{
       hao = 0;
       make_root();
       for(int i=1;i<=500100;i++)
       {
           father[i]=i;
       }
       memset(du,0,sizeof(du));
      // int numm = 0;
    while(scanf("%s%s",a,b)!=EOF)
       {
              int u = INsert(a);
              int v = INsert(b);
              du[u]++;
              du[v]++;
              merge(u,v);
            //  numm++;
             // if(numm==5)
              //  break;
       }
       ans = 0;
       oula();
       if(( ans == 0|| ans ==2)&& !flag )
        printf("Possible\n");
       else
       printf("Impossible\n");
       return 0;
}