poj 2513 欧拉图/trie

Posted 2020-10-07 zzq

http://poj.org/problem?id=2513

 

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 37812		Accepted: 9907

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.

Source

The UofA Local 2000.10.14

   对每一对颜色进行连接一条无向边然后跑欧拉图验证是否可行，用了Trie来哈希字符串，并查集判断是否是连通图，坑点是有空串，，，所以可能有0个集合，此时输出"Possible";

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<map>
 5 using namespace std;
 6 int N;
 7 struct node
 8 {
 9     node *next[26];
10     int v;
11     node(){v=0;memset(next,NULL,sizeof(next));}
12 }*root;
13 int gec(char *s)
14 {
15     int len=strlen(s);
16     node *p=root;
17     for(int i=0;i<len;++i)
18     {
19         if(p->next[s[i]-‘a‘]==NULL) p->next[s[i]-‘a‘]=new node();
20         p=p->next[s[i]-‘a‘];
21     }
22     if(p->v==0) p->v=++N;
23     return p->v;
24 }
25 int cnt[500005];
26 int f[500005];
27 int getf(int v){return f[v]==v?v:f[v]=getf(f[v]);}
28 int main()
29 {
30     char s1[15],s2[15];
31     int i;
32     for(i=1;i<=500000;++i)f[i]=i;
33     root=new node();
34     while(scanf("%s%s",s1,s2)!=EOF){
35         int u=gec(s1);
36         int v=gec(s2);
37         int fu=getf(u),fv=getf(v);
38         if(fu!=fv) f[fv]=fu;
39         cnt[u]++;
40         cnt[v]++;
41     }
42     int s=0,x=0;
43     for(i=1;i<=N;++i)
44     {
45         if(cnt[i]%2==1) s++;
46         if(i==getf(i))x++;
47     }
48     if(x<=1&&(s==0||s==2)) puts("Possible");
49     else puts("Impossible");
50     return 0;
51 }