poj 2513 Colored Sticks

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题目链接：点击打开链接

Description

You are given a bunch(群) of wooden sticks. Each endpoint(端点) of each stick is colored with some color. Is it possible to align(结盟) the sticks in a straight line such that the colors of the endpoints that touch are of the same color?

Input

Input(投入) is a sequence(序列) 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(小写字母) letters no longer than 10 characters. There is no more than 250000 sticks.

Output

If the sticks can be aligned(结盟) in the desired way, output(输出) a single line saying Possible, otherwise output Impossible.

Sample Input

blue redred violetcyan blueblue magentamagenta cyan

Sample Output

Possible

题意：给出多对颜色，逆可以把同种颜色想象成一个数字，问能不能构成欧拉回路．

ps：先用字典树把这些字符串存一下，用字典树在搜索某个点是否存在是速度会快．将这个颜色的度++，

在用并查集将这个颜色对串起来．

最后判断一下是否构成欧拉回路：

欧拉回路的定义：在一条联通分量上，且奇数度的点为0或2个

<span style="font-size:24px;">//字典树＋并查集＋欧拉回路#include <iostream>#include<cstring>#include<cstdio>#include<cstdlib>using namespace std;int arr[520000],deg[520000];struct node{    int x;    struct node *next[27];}*root;int top;struct node *creat(){    struct node *p;    p=new node;    p->x=0;    for(int i=0; i<27; i++)    {        p->next[i]=NULL;    }    return p;}int search1(struct node *root,char s[]){    int len=strlen(s);    int k;    struct node *p=root;    for(int i=0; i<len; i++)    {        k=s[i]-'a';        if(p->next[k]==NULL)        {            p->next[k]=creat();        }        p=p->next[k];    }    if(p->x==0)    {        p->x=top++;    }    return p->x;}int f(int x){   while(x!=arr[x])   {       x=arr[x];   }   return x;}int main(){    int i,k1,k2;    memset(deg,0,sizeof(deg));    root=creat();    for(i=0; i<=520000; i++)    {        arr[i]=i;    }    top=1;    char s1[15],s2[15];    while(~scanf("%s%s",s1,s2))    {        k1=search1(root,s1);        k2=search1(root,s2);        deg[k1]++;        deg[k2]++;        if(f(k1)!=f(k2))        {            arr[f(k1)]=f(k2);        }    }    int k=f(1);    for( i=2; i<top; i++)    {        if(k!=f(i))        {            break;        }    }    if(i<top)    {        printf("Impossible\n");    }    else    {        int ans=0;        for(int j=1; j<top; j++)        {            if(deg[j]%2==1)ans++;        }        if(ans==0||ans==2)        {            printf("Possible\n");        }        else printf("Impossible\n");    }    return 0;}</span>

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