zoukankan      html  css  js  c++  java
  • 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 }
  • 相关阅读:
    JavaScript语法规范
    js中的逻辑与(&&)和逻辑或(||)
    canvas边界与摩擦力
    canvas模拟重力效果
    MyEclispe 2015 CI 15发布(附下载)
    MyEclipse 2015 CI 14发布(附下载)
    10本Java经典书目推荐
    MyEclipse 2015免费在线公开课,2月5日开讲
    DevExpress应用案例--语音识别器Dual Writer
    DevExtreme官方视频教程分享
  • 原文地址:https://www.cnblogs.com/CKboss/p/3315007.html
Copyright © 2011-2022 走看看