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:
    [每天进步一点 流水账]第4周
    单指令流多数据流( SIMD)
    [每天进步一点 流水账]第2周
    写时复制技术(COW)
    ECMAScript 运算符乘性运算符
    ECMAScript 运算符Boolean 运算符
    ECMAScript 基础保留字
    ECMAScript 基础关键字
    ECMAScript 运算符一元运算符
    ECMAScript 基础原始类型
  • 原文地址:https://www.cnblogs.com/asuran/p/1579981.html
Copyright © 2011-2022 走看看