本文主要是介绍POJ 2513 Colored Sticks(字典树+欧拉路径)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=11158
Colored Sticks
| Time Limit: 5000MS | Memory Limit: 128000KB | 64bit IO Format: %I64d & %I64u |
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.
相关知识点：
欧拉回路和欧拉路径的判断
欧拉回路：
无向图：每个顶点的度数都是偶数，则存在欧拉回路。
有向图：每个顶点的入度都等于出度，则存在欧拉回路。
欧拉路径：
无向图：当且仅当该图所有顶点的度数为偶数 或者 除了两个度数为奇数外其余的全是偶数。
有向图：当且仅当该图所有顶点 出度=入度 或者 一个顶点 出度=入度+1，另一个顶点 入度=出度+1，其他顶点 出度=入度。
本题和hdu 1116（上一篇博客）稍有不同，这是无向图，那个是有向图（单词不可能倒着写）。判定条件不要混淆。相关博客： http://blog.csdn.net/zzran/article/details/9106329
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=5e5+5;
struct node{int num;bool is_word;node *next[26];node(){num=is_word=0;memset(next,NULL,sizeof(next));}
};
node *root;
int n,c[maxn],f[maxn];
void insert(char s[]){node *p=root;for(int i=0;s[i];i++){int dex=s[i]-'a';if(p->next[dex]==0) p->next[dex]=new node();p=p->next[dex];}if(p->is_word==0){ // important! because of num++p->num=++n; //num is dex of stringp->is_word=1;}
}
int find(int x){if(x==f[x]) return x;f[x]=find(f[x]);return f[x];
}
int getdex(char s[]){node *p=root;for(int i=0;s[i];p=p->next[s[i]-'a'],i++); // not need lengthreturn p->num;
}
int main()
{//freopen("cin.txt","r",stdin);n=0;memset(c,0,sizeof(c));memset(f,0,sizeof(f));root=new node();char a[15],b[15];int adex,bdex;while(~scanf("%s%s",a,b)){insert(a);insert(b);adex=getdex(a);bdex=getdex(b);c[adex]++; //all countc[bdex]++;if(!f[adex]) f[adex]=adex; // U gather。if(!f[bdex]) f[bdex]=bdex;f[find(adex)]=f[find(bdex)];}int oddsum=0;bool judge=1;for(int i=1;i<=n;i++){if(c[i]&1) oddsum++;if(find(i)!=f[1]){judge=0;break;}}if((oddsum==0 || oddsum==2)&&judge) puts("Possible");else puts("Impossible");return 0;
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