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；

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 int degree[500010],set[500010],id = 1;

 5 

 6 struct node

 7 {

 8     int flag;

 9     int id;

10     struct node* next[26];

11 };

12 struct node* root;

13 

14 //开辟新结点

15 struct node* creat()

16 {

17     struct node *p = (struct node*)malloc(sizeof(struct node));

18     p->flag = 0;

19     for(int i = 0; i < 26; i++)

20         p->next[i] = NULL;

21     return p;

22 }

23 

24 int find(int x)

25 {

26     if(set[x] != x)

27         set[x] = find(set[x]);//路径压缩；

28     return set[x];

29 }

30 

31 //字典树哈希

32 int Hash(char s[])

33 {

34     struct node *p = root;

35     for(int i = 0; s[i]; i++)

36     {

37         if(p->next[s[i]-'a'] == NULL)

38             p->next[s[i]-'a'] = creat();

39         p = p->next[s[i]-'a'];

40     }

41     if(p->flag != 1)

42     {

43         p->flag = 1;

44         p->id = id++;

45     }

46     return p->id;

47 }

48 

49 int check()

50 {

51     int sum = 0;

52     int x = find(1);

53     for(int i = 2; i < id; i++)

54         if(find(i) != x)//没有共同祖先，图是不连通的，直接返回；

55             return 0;

56     for(int i = 1; i < id; i++)

57     {

58         if(degree[i]%2)

59             sum++;

60     }

61     if(sum == 0 || sum == 2)

62         return 1;//图是连通的并且奇度数是0或2，说明有欧拉路；

63     return 0;//图是连通的但奇度数不是0或2也不存在欧拉路；

64 }

65 

66 int main()

67 {

68     memset(degree,0,sizeof(degree));

69     for(int i = 1; i <= 500000; i++)

70         set[i] = i;//所有节点初始化为一棵树

71     char s1[12],s2[12];

72     int u,v;

73     root = creat();

74     while(scanf("%s %s",s1,s2) != EOF)

75     {

76         u = Hash(s1);

77         v = Hash(s2);

78         degree[u]++;

79         degree[v]++;

80         int x = find(u);

81         int y = find(v);

82         if(x != y)

83             set[x] = y;

84     }

85     if(check()) printf("Possible\n");

86     else printf("Impossible\n");

87     return 0;

88 }

View Code