zoukankan      html  css  js  c++  java
  • POJ 2513, Colored Sticks

    Time Limit: 5000MS  Memory Limit: 128000K
    Total Submissions: 13696  Accepted: 3384


    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


    // POJ2513.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    using namespace std;

    //Tries
    struct Trie
    {
        Trie():id(
    -1),end(false){memset(next, 0sizeof(next));}
        
    int id;
        
    bool end;
        Trie
    * next[26];
    };
    static int num = 0;
    int GetID(char *x, Trie* root)
    {
        Trie 
    *temp = root;
        
    for (int i = 0; i < strlen(x); ++i)
        {
            
    if (temp->next[x[i]-'a'== NULL)
                temp
    ->next[x[i]-'a'= new Trie;
            temp 
    = temp->next[x[i]-'a'];
        }
        
    if (temp->end)return temp->id;
        temp
    ->end = true;
        temp
    ->id = num++;
        
    return temp->id;
    }

    //Disjoint set
    int Find(int x, int f[])
    {
        
    if(x != f[x])
            
    return f[x] = Find(f[x], f);
        
    return f[x];
    };
    void Union(int x,int y, int f[])
    {
        f[Find(x, f)] 
    = f[Find(y, f)];
    };

    int main(int argc, char* argv[])
    {
        Trie root;
        
    char w[30], w1[12], w2[12];
        
    int degree[500001];
        memset(degree,
    0,sizeof(degree));
        
    int f[500001];
        
    for(int i = 0; i < 500001++i) f[i] = i;
        
        
    //create Trie tree
        while (gets(w) && w[0!= 0)
        {
            sscanf(w,
    "%s %s", w1, w2);
            
    int x = GetID(w1, &root);
            
    int y = GetID(w2, &root);
            
    ++degree[x];
            
    ++degree[y];
            Union(x,y,f);
        }

        
    //check odd points
        int odd = 0;
        
    for (int i = 0; i < num; ++i)
            
    if (degree[i] & 1 != 0++odd;

        
    //check connected
        int x = Find(1,f);
        
    for (int i = 0; i < num ; ++i)
            
    if (x != Find(i,f))
            {
                odd 
    = -1;
                
    break;
            };    

        
    if (odd == 0 || odd == 2) cout << "Possible" << endl;
        
    else cout << "Impossible" << endl;
        
    return 0;
    }

  • 相关阅读:
    高性能网络编程2----TCP消息的发送
    高性能网络编程1----accept建立连接
    Android之怎样使用ListView列表视图
    创建hive整合hbase的表总结
    最新版本号cocos2d&#173;2.0&#173;x&#173;2.0.2使用新资源载入策略!不再沿用-hd、-
    在NSUserDefaults中保存自己定义的对象
    Light oj 1138
    一个NHibernate的BUG
    hbase exporter importer 导出 导入
    Gulp帮你自己主动搞定coffee和scss的compile
  • 原文地址:https://www.cnblogs.com/asuran/p/1579981.html
Copyright © 2011-2022 走看看