Poj 2513 colored sticks and query the Euler's loop judgment of the Set dictionary tree

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Colored sticks

Time limit:5000 Ms		Memory limit:128000 K
Total submissions:30273		Accepted:8002

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 redred violetcyan blueblue magentamagenta cyan

Sample output

Possible

The two ends of a stick can be connected in the same color.

Ask if all the sticks can be connected together.

#include<cstdio>#include<cstring>#include<iostream>#include<cstdlib>#include<algorithm>using namespace std;struct node{    int next[26];    int id;}trie[250010];int head;int father[25001    0];int du[250010];int ids=1;int tree_add(char *str){    int s=0;    int len=strlen(str);    int i;    int w;    for(i=0;i<len;i++)    {        w=str[i]-'a';        if(trie[s].next[w]==-1)        {            trie[s].next[w]=++head;        }        s=trie[s].next[w];    }    if(trie[s].id==0)    {        trie[s].id=ids++;        return trie[s].id;    }    else        return trie[s].id;}int find(int x){    if(x==father[x])        return x;    int t=father[x];    father[x]=find(father[x]);    return father[x];}int main(){    char str1[21];    char str2[11];    int i;    int a,b;    head=0;    int j;    for(i=0;i<=250010;i++)    {        trie[i].id=0;        for(j=0;j<26;j++)        {            trie[i].next[j]=-1;        }        father[i]=i;        du[i]=0;    }    while(gets(str1)!=NULL&&strcmp(str1,"")!=0)    {        for(i=0;str1[i]!='\0';i++)        {            if(str1[i-1]==' '&&str1[i]!=' ')            {                strcpy(str2,str1+i);                break;            }        }        for(i=0;str1[i]!='\0';i++)        {            if(str1[i]==' ')            {                str1[i]='\0';                break;            }        }        a=tree_add(str1);        b=tree_add(str2);        du[a]++;        du[b]++;       // printf("a = %d b = %d \n",a,b);        int x=find(a);        int y=find(b);        if(x!=y)        {            father[x]=y;        }    }    int aaa=0;    for(i=1;i<=ids-1;i++)    {        if(father[i]==i)        {            aaa++;        }        if(aaa>1)        {            printf("Impossible\n");            return 0;        }    }    aaa=0;    for(i=1;i<=ids-1;i++)    {        if(du[i]%2==1)        {            aaa++;        }    }  //  printf("%d \n",aaa);    if(aaa>2||aaa==1)    {        printf("Impossible\n");    }    else    {        printf("Possible\n");    }    return 0;}