POJ 2513 Colored Sticks



Colored Sticks
Time Limit: 5000MS   Memory Limit: 128000K
Total Submissions: 28036   Accepted: 7428

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 

无向图欧拉通路：连通图+两个点的度数是奇数。。。。。

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 
  5 using namespace std;
  6 
  7 const int maxnode=10000000;
  8 int worldnum=0,Father[510000],USize[510000],degree[510000];
  9 
 10 struct Trie
 11 {
 12     int tot,root,child[maxnode][26];
 13     int flag[maxnode];
 14     Trie()
 15     {
 16         memset(child[1],0,sizeof(child[1]));
 17         memset(flag,-1,sizeof(flag));
 18         root=tot=1;
 19     }
 20     void Insert(const char* str)
 21     {
 22         int *cur=&root;
 23         for(const char* p=str;*p;*p++)
 24         {
 25             cur=&child[*cur][*p-'a'];
 26             if(*cur==0)
 27             {
 28                 *cur=tot++;
 29                 memset(child[tot],0,sizeof(child[tot]));
 30                 flag[tot]=-1;
 31             }
 32         }
 33         flag[*cur]=worldnum++;
 34     }
 35 
 36     int Query(const char*str)
 37     {
 38         int* cur=&root;
 39         for(const char* p=str;*p&&*cur;p++)
 40         {
 41             cur=&child[*cur][*p-'a'];
 42         }
 43         if(*cur&&~flag[*cur])
 44         {
 45             return flag[*cur];
 46         }
 47         else
 48         {
 49             Insert(str);
 50             return worldnum-1;
 51         }
 52     }
 53 }T;
 54 
 55 int Findfather(int x)
 56 {
 57     if(Father[x]==x)
 58     {
 59         return x;
 60     }
 61     else
 62     {
 63         Father[x]=Father[Father[x]];
 64         return Findfather(Father[x]);
 65     }
 66 }
 67 
 68 void Disjoin()
 69 {
 70     for(int i=0;i<510000;i++)
 71     {
 72         Father[i]=i; USize[i]=1;
 73     }
 74 }
 75 
 76 void Unionset(int a,int b)
 77 {
 78     int Fa=Findfather(a);
 79     int Fb=Findfather(b);
 80     if(Fa==Fb) return ;
 81     if(USize[Fa]>=USize[Fb])
 82     {
 83         Father[Fa]=Fb;
 84         USize[Fb]+=USize[Fa];
 85     }
 86     else
 87     {
 88         Father[Fb]=Fa;
 89         USize[Fa]+=USize[Fb];
 90     }
 91 }
 92 
 93 int main()
 94 {
 95     char str1[15],str2[15];
 96     Disjoin();
 97     while(scanf("%s %s",str1,str2)!=EOF)
 98     {
 99         int a=T.Query(str1);
100         int b=T.Query(str2);
101         degree[a]++; degree[b]++;
102         Unionset(a,b);
103     }
104     int sum=0;
105     for(int i=0;i<worldnum;i++)
106     {
107         if(degree[i]%2==1) sum++;
108     }
109     if(sum>2)  { puts("Impossible"); return 0;}
110     for(int i=1;i<worldnum;i++)
111     {
112         if(Findfather(0)!=Findfather(i))
113             { puts("Impossible"); return 0;}
114     }
115     puts("Possible");
116     return 0;
117 }

 

posted @ 2013-09-11 17:22  码代码的猿猿  阅读(236)  评论(0)  编辑  收藏  举报