poj2513 연결 성냥 사전 트리+오라 통로 좋은 문제

Colored Sticks
Time Limit: 5000MS
 
Memory Limit: 128000K
Total Submissions: 27134
 
Accepted: 7186
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
 
제목: 여러 개의 화채가 있는데 머리마다 색깔이 같은 색깔이 있어서 다른 성냥과 연결할 수 있는지 물어보세요.
 
사고방식: 분명히 이것은 오로라 통로 문제이고 그림을 그리는 완전한 통로이다
사전 트리로 문자열에 대응하는 id 표시하기
맵으로 시간 초과

#include<stdio.h>

#include<string>

#include<string.h>

#include<map>

#include<malloc.h>

using namespace std;

#define N 250000+10

char s1[100],s2[100];

int rank[N],fa[N],num[N],co=0;

struct haha

{

    int id;

    struct haha *next[26];

}*root;

struct haha * creat()

{

    int i;

    struct haha *p;

    p=(struct haha *)malloc(sizeof(struct haha));

    p->id=-1;

    for(i=0;i<26;i++) p->next[i]=NULL;

    return p;

};

int update(char *s)

{

     int d,pos,i;

     struct haha *p;

     p=root;d=strlen(s);

     for(i=0;i<d;i++)

     {

         pos=s[i]-'a';

         if(p->next[pos]==NULL)

         {

             p->next[pos]=creat();

             p=p->next[pos];

         }

         else

         {

             p=p->next[pos];

         }

     }

     if(p->id==-1)

     p->id=co++;

     return p->id;

}

int  find(int n)

{

    return n==fa[n]?n:fa[n]=find(fa[n]);

}

void join(int a,int b)

{

        if(rank[a]>rank[b])

        {

            fa[b]=a;

            rank[a]+=rank[b];

        }

        else

        {

            fa[a]=b;

            rank[b]+=rank[a];

        }

}

int main()

{

    int i,j,k,cnt=0,a,b,rem;

    for(i=0;i<N;i++)

    {

        rank[i]=1;fa[i]=i;

    }

    memset(num,0,sizeof(num));

    root=creat();

    while(scanf("%s %s",s1,s2)!=EOF)

    {

         a=update(s1);

         b=update(s2);



         num[a]++;num[b]++;

         a=find(a);b=find(b);

         if(a==b) continue;

          join(a,b);

    }

    cnt=0;

    int temp=find(0);

    for(i=0;i<N;i++)

    {

        if(num[i]%2==1) cnt++;

        if(cnt>2) {printf("Impossible
");return 0;}

        if(num[i]!=0&&find(i)!=temp) {printf("Impossible
");return 0;}

    }

    printf("Possible
");

    return 0;

}