Colored Sticks
| Time Limit: 5000MS | Memory Limit: 128000K | |
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
回顾了一下欧拉图的知识。。整个图的奇度点个数如果>=3或只有1个 就不能一笔画画完
中间re了几次。。因为trie里我开了n*30，但是外面数组只开了n...
1 #include<set> 2 #include<queue> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 const int N = 250010; 10 #define For(i,n) for(int i=1;i<=n;i++) 11 #define For0(i,n) for(int i=0;i<n;i++) 12 #define Rep(i,l,r) for(int i=l;i<=r;i++) 13 14 char s1[11],s2[11]; 15 int n,in[N*30],fa[N*30],ans; 16 struct Trie{ 17 int sz,ch[N*30][26],v[N*30],Loc[N*30]; 18 Trie(){sz=1;} 19 int insert(char st[]){ 20 int now=1;v[1]++;int len=strlen(st); 21 For0(i,len){ 22 int id=st[i]-'a'; 23 if(!ch[now][id]) ch[now][id]=++sz; 24 v[now=ch[now][id]]++; 25 } 26 if(Loc[now]) return Loc[now]; 27 else return Loc[now]=++n; 28 } 29 }trie; 30 31 int find(int i){ 32 if(fa[i]==i) return i; 33 else return fa[i]=find(fa[i]); 34 } 35 36 int main(){ 37 #ifndef ONLINE_JUDGE 38 freopen("trie.in","r",stdin); 39 #endif // ONLINE_JUDGE 40 For0(i,N*30) fa[i]=i; 41 while(scanf("%s %s",&s1,&s2)!=EOF){ 42 int x=trie.insert(s1),y=trie.insert(s2); 43 in[x]++;in[y]++; 44 int fx=find(x),fy=find(y); 45 if(fx!=fy) fa[fx]=fy; 46 } 47 int checks=find(1); 48 Rep(i,2,n) if(find(i)!=checks){ 49 puts("Impossible "); 50 return 0; 51 } 52 For(i,n) { 53 in[i]=in[i]%2; 54 if(in[i]) ans++; 55 } 56 if(ans==1||ans>=3) puts("Impossible"); 57 else puts("Possible"); 58 return 0; 59 }