Colored Sticks (字典树哈希+并查集+欧拉路)

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 27704		Accepted: 7336

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.

 

题意：给定若干个棒，棒的两端涂上不同的颜色，问是否能将棒首尾相连且不同棒相接的一端颜色相同；

 

思路：题意容易理解，但开始完全没思路，经大神指点后，可以用欧拉路的思想；可以把涂颜色的棒的两端看成结点，

         把木棒看成边，相同颜色的就是一个结点，要将木棒连成一个直线，也就是“一笔画”问题；

         无向图存在欧拉路的充要条件是：

         >图是连通的（可以用并查集判断，开始将每个点初始化一棵树，经过输入将有相同祖先的结点合并到一个集合中，

           最后任意枚举一个节点，若他们有共同的祖先，说明图是连通的）；

         >度数为奇数的结点有0个或两个；

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int degree[500010],set[500010],id = 1;

struct node
{
    int flag;
    int id;
    struct node* next[26];
};
struct node* root;

//开辟新结点
struct node* creat()
{
    struct node *p = (struct node*)malloc(sizeof(struct node));
    p->flag = 0;
    for(int i = 0; i < 26; i++)
        p->next[i] = NULL;
    return p;
}

int find(int x)
{
    if(set[x] != x)
        set[x] = find(set[x]);//路径压缩；
    return set[x];
}

//字典树哈希
int Hash(char s[])
{
    struct node *p = root;
    for(int i = 0; s[i]; i++)
    {
        if(p->next[s[i]-'a'] == NULL)
            p->next[s[i]-'a'] = creat();
        p = p->next[s[i]-'a'];
    }
    if(p->flag != 1)
    {
        p->flag = 1;
        p->id = id++;
    }
    return p->id;
}

int check()
{
    int sum = 0;
    int x = find(1);
    for(int i = 2; i < id; i++)
        if(find(i) != x)//没有共同祖先，图是不连通的，直接返回；
            return 0;
    for(int i = 1; i < id; i++)
    {
        if(degree[i]%2)
            sum++;
    }
    if(sum == 0 || sum == 2)
        return 1;//图是连通的并且奇度数是0或2，说明有欧拉路；
    return 0;//图是连通的但奇度数不是0或2也不存在欧拉路；
}

int main()
{
    memset(degree,0,sizeof(degree));
    for(int i = 1; i <= 500000; i++)
        set[i] = i;//所有节点初始化为一棵树
    char s1[12],s2[12];
    int u,v;
    root = creat();
    while(scanf("%s %s",s1,s2) != EOF)
    {
        u = Hash(s1);
        v = Hash(s2);
        degree[u]++;
        degree[v]++;
        int x = find(u);
        int y = find(v);
        if(x != y)
            set[x] = y;
    }
    if(check()) printf("Possible
");
    else printf("Impossible
");
    return 0;
}