Colored Sticks (字典树哈希+并查集+欧拉路)

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 27704		Accepted: 7336

Description

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

Hint

Huge input,scanf is recommended.

 

题意：给定若干个棒，棒的两端涂上不同的颜色，问是否能将棒首尾相连且不同棒相接的一端颜色相同；

 

思路：题意容易理解，但开始完全没思路，经大神指点后，可以用欧拉路的思想；可以把涂颜色的棒的两端看成结点，

         把木棒看成边，相同颜色的就是一个结点，要将木棒连成一个直线，也就是“一笔画”问题；

         无向图存在欧拉路的充要条件是：

         >图是连通的（可以用并查集判断，开始将每个点初始化一棵树，经过输入将有相同祖先的结点合并到一个集合中，

           最后任意枚举一个节点，若他们有共同的祖先，说明图是连通的）；

         >度数为奇数的结点有0个或两个；

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 int degree[],set[],id = ;

 struct node
 {
     int flag;
     int id;
     struct node* next[];
 };
 struct node* root;

 //开辟新结点
 struct node* creat()
 {
     struct node *p = (struct node*)malloc(sizeof(struct node));
     p->flag = ;
     for(int i = ; i < ; i++)
         p->next[i] = NULL;
     return p;
 }

 int find(int x)
 {
     if(set[x] != x)
         set[x] = find(set[x]);//路径压缩；
     return set[x];
 }

 //字典树哈希
 int Hash(char s[])
 {
     struct node *p = root;
     for(int i = ; s[i]; i++)
     {
         if(p->next[s[i]-'a'] == NULL)
             p->next[s[i]-'a'] = creat();
         p = p->next[s[i]-'a'];
     }
     if(p->flag != )
     {
         p->flag = ;
         p->id = id++;
     }
     return p->id;
 }

 int check()
 {
     int sum = ;
     int x = find();
     for(int i = ; i < id; i++)
         if(find(i) != x)//没有共同祖先，图是不连通的，直接返回；
             return ;
     for(int i = ; i < id; i++)
     {
         if(degree[i]%)
             sum++;
     }
     if(sum ==  || sum == )
         return ;//图是连通的并且奇度数是0或2，说明有欧拉路；
     return ;//图是连通的但奇度数不是0或2也不存在欧拉路；
 }

 int main()
 {
     memset(degree,,sizeof(degree));
     for(int i = ; i <= ; i++)
         set[i] = i;//所有节点初始化为一棵树
     char s1[],s2[];
     int u,v;
     root = creat();
     while(scanf("%s %s",s1,s2) != EOF)
     {
         u = Hash(s1);
         v = Hash(s2);
         degree[u]++;
         degree[v]++;
         int x = find(u);
         int y = find(v);
         if(x != y)
             set[x] = y;
     }
     if(check()) printf("Possible\n");
     else printf("Impossible\n");
     return ;
 }