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poj 2513 Colored Sticks

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Colored Sticks

Time Limit:

5000MS
Memory Limit:

128000K

Total Submissions:

35734
Accepted:

9358

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.
/*

就是给n根木棒，木棒两端都有颜色，判断能不能把木棒合并成一条，接触的木棒端必须是相同颜色的。


给定一些木棒，木棒两端都涂上颜色，求是否能将木棒首尾相接，连成一条直线，要求不同木棒相接的一边必须是相同颜色的。

*/

用欧拉路来做，可以把木棒的两段看作是两个点，然后我们就可以放开思想做题了
我们知道无向图的欧拉路要满足两个特性1. 所有点的边数是奇数个的有0个或2个，2. 这个图要是联通的
这样就好办了，我们把每一个木棒的两段看成是两个点的边，然后就会得到点的边数了，当然一样的字母是同一个点，在然后我们就要用并查集的知识来看看图是否为联通图，
这里有一个难点就是我们必须用字典树的知识查找相同的字母，字典树我在以前写到的，
这个题目的代码如下： 
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

int father[510005];
int deg[510005];
int cnt, ant;
typedef struct node
{
    int x;
    node *next[26];
}*Trie;

void init()
{
    int i;
    for ( i = 1;i < 510005; i++ )
    {
        father[i] = i;
    }
    memset(deg, 0, sizeof(deg));
    ant = 0;
}

int search1(Trie root, char *s)
{
    int k;
    while ( *s )
    {
        if ( *(s+1) )
        {
            k = *s-'a';
            if ( root->next[k] == NULL )
                return 0;
            root = root->next[k];
            s++;
        }
        else
        {
            k = *s-'a';
            if ( root->next[k] == NULL )
                return 0;
            if ( root->next[k]->x != -1 )
                return root->next[k]->x;
            return 0;
        }
    }
    return 0;
}

void insert(Trie root, char *s)
{
    int i, k;
    while ( *s )
    {
        k = *s-'a';
        if ( root->next[k] == NULL )
        {
            root->next[k] = new node;
            if ( !*(s+1) )
                root->next[k]->x = ++ant;
            else
                 root->next[k]->x = -1;
            for ( i = 0;i < 26; i++ )
            {
                root->next[k]->next[i] = NULL;
            }
        }
        else
        {
            if ( !*(s+1) )
                root->next[k]->x = ++ant;
        }
        root = root->next[k];
        s++;
    }
}

int find(int x)
{
    while ( x != father[x] )
    {
        x = father[x];
    }
    return x;
}

void meget(int x, int y)
{
    int fx = find(x);
    int fy = find(y);
    if ( fx != fy )
        father[fx] = fy;
}

int euler()
{
    int i;
    int cnt = 0;
    for ( i = 1;i <= ant; i++ )
    {
        if ( deg[i]%2 != 0 )
            cnt++;
    }
    if ( cnt != 0&&cnt != 2 )
        return 0;
    int g = find(1);
    for ( i = 2;i <= ant; i++ )
    {
        if ( g != find(i) )
            return 0;
    }
    return 1;
}

int main()
{
    int x, y;
    int i;
    char s1[15], s2[15];
    init();
    Trie root;
    root = new node;
    for ( i = 0;i < 26; i++ )
    {
        root->next[i] = NULL;
    }
    while ( ~scanf ( "%s %s", s1, s2) )
    {
        x = search1( root, s1 );
        y = search1( root, s2 );
        if ( x == 0 )
        {
            insert( root, s1 );
            x = ant;
        }
        if ( y == 0 )
        {
            insert( root, s2 );
            y = ant;
        }
        deg[x]++;
        deg[y]++;
        meget( x, y );
    }
    if ( euler() )
        printf ( "Possible\n" );
    else
        printf ( "Impossible\n" );
}

代码菜鸟，如有错误请多包涵！！！

如有帮助记得支持我一下，谢谢！！！


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