poj 2513 连接火柴 字典树+欧拉通路 好题

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

 

题意： 有很多个火彩 每头都有颜色   相同的颜色可以和另外的火柴相接  问这些火柴能不能连接成一条线 

 

思路：  很明显这是一个欧拉通路问题      一笔画画完整个通路 

用字典树标记字符串对应的id

用map超时

#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;
}