POJ 2513 Colored Sticks 欧拉回路+trie树

题目

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

即木棍相接的颜色得相同~

本来想用哈希处理，WA了，提供从建议改用trie树，现在看来练习中难的够呛的算法也只不过是为了其他算法做配角

不仅仅说是并查集检验欧拉回路==trie树不也是嘛

#include <iostream>#include<cstdio>#include<cstring>using namespace std;int total;class trie{public:    trie* next[26];    char *color;    int value; //标记这里是不是一个单词    trie()    {        for(int i=0;i<26;i++)            next[i]=0;        color=NULL;        value=-1;    }}root;int insert(trie *p,char* s){    int k=0;    while(s[k]!='\0')    {        if(!p->next[s[k]-'a'])            p->next[s[k]-'a']=new trie;        p=p->next[s[k]-'a'];        k++;    }    if(p->value==-1)    p->value=(total++);    return p->value;    //在这里可根据具体题意来加东西}int a[500019];int total1[500010],total2[500010];int find(int x){    if(x!=a[x]) a[x]=find(a[x]);    return a[x];}void uni(int x,int y){    x=find(x),y=find(y);    if(x!=y) a[x]=y;}int main(){    //freopen("cin.txt","r",stdin);    int t,n,sum;    char str1[20],str2[20];    trie *p=&root;           // scanf("%d",&n);          // memset(vis,false,sizeof(vis));            for(int i=0;i<=500010;i++) a[i]=i;            sum=0,total=0;            int in=0,out=0;            memset(total1,0,sizeof(total1));            memset(total2,0,sizeof(total2));            while(~scanf("%s%s",str1,str2))            {                int num1=insert(p,str1);                int num2=insert(p,str2);                total1[num1]++;//total2[num2]++;                total1[num2]++;                uni(num1,num2);             //   printf("%d %d\n",num1,num2);               // if(!vis[num1]){ total++;vis[num1]=true;}               // if(!vis[num2]){ total++;vis[num2]=true;}               //total=max(total,max(num1,num2));            }            for(int i=0;i<total;i++)            {                if(i==a[i]) sum++;                /*if(total1[i]!=total2[i])                {                    if(total1[i]==total2[i]+1) in++;//cout<<i<<endl;                    else if(total1[i]==total2[i]-1) out++;                    else {sum+=100;break;}                }*/                if(total1[i]&1) in++;                if(sum>1) break;                if(in>2) break;            }           // if((sum==0||sum==1)&&((in==1&&out==1)||(in==0&&out==0)))printf("Possible\n");             if((sum==0||sum==1)&&(in==0||in==2))                  printf("Possible\n");              else printf("Impossible\n");          //  printf("sum=%d  in=%d  out=%d\n",sum,in,out);    return 0;}