POJ 2513, Colored Sticks

最新推荐文章于 2019-09-19 15:40:13 发布

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最新推荐文章于 2019-09-19 15:40:13 发布

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原文链接：http://www.cnblogs.com/asuran/archive/2009/10/10/1579981.html

本文详细解析了一种算法，该算法用于判断是否可以将一束两端涂有颜色的木棍排成一行，使得相邻木棍接触的两端颜色相同。通过构建字典树和并查集数据结构，算法有效地解决了这一问题。

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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, 0, sizeof(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;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/10/1579981.html