题目链接：
http://poj.org/problem?id=2513
http://bailian.openjudge.cn/practice/2513?lang=en_US
Time Limit: 5000MS Memory Limit: 128000K
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.
题意：
给出若干木棒，每个木棒两个端点有两个颜色，两个木棒的端点颜色相同，则可以接在一起。问是否把所有木棒接成一根。
题解：
将每个颜色都看成一个节点，木棒就能看成一条连接两个端点的无向边。问题就变成在给定一个无向图问是否存在欧拉路。
欧拉路存在的充要条件是：① 图是连通的； ② 所有节点的度为偶数，或者有仅有两个度数为奇数的节点。
这道题，求顶点的度数很好求，但是怎么判断图是否连通呢？通常来说，DFS/BFS或者并查集都能做。
不过，本题字典树的用处就是代替map，将输入字符串快速hash成一个值。
AC代码（BFS判图的连通性）：
#include<bits/stdc++.h>
using namespace std;
const int maxn=250000+10;
int degree[2*maxn];
vector<int> G[2*maxn];
namespace Trie
{
const int SIZE=maxn*10;
int sz;
int idtot;
struct TrieNode{
int ed;
int nxt[26];
}trie[SIZE];
void init()
{
idtot=0;
sz=1;
}
int insert(const string& s)
{
int p=1;
for(int i=0;i<s.size();i++)
{
int ch=s[i]-'a';
if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
p=trie[p].nxt[ch];
}
if(!trie[p].ed) trie[p].ed=++idtot;
return trie[p].ed;
}
};
bool vis[2*maxn];
int bfs(int s)
{
int res=0;
memset(vis,0,sizeof(vis));
queue<int> Q;
Q.push(s), vis[s]=1, res++;
while(!Q.empty())
{
int u=Q.front(); Q.pop();
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(!vis[v]) Q.push(v), vis[v]=1, res++;
}
}
return res;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
string s1,s2;
Trie::init();
while(cin>>s1>>s2)
{
int u=Trie::insert(s1), v=Trie::insert(s2);
G[u].push_back(v);
G[v].push_back(u);
degree[u]++;
degree[v]++;
}
int cnt=0;
for(int i=1;i<=Trie::idtot;i++) {
if(degree[i]%2) cnt++;
}
if(cnt==1 || cnt>2) cout<<"Impossible\n";
else
{
if(bfs(1)<Trie::idtot) cout<<"Impossible\n";
else cout<<"Possible\n";
}
}
AC代码（并查集判图的连通性）：
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=250000+5;
namespace Trie
{
const int SIZE=maxn*10;
int sz;
int idcnt;
struct TrieNode{
int ed;
int nxt[26];
}trie[SIZE];
void init()
{
idcnt=0;
sz=1;
}
int insert(char *s)
{
int len=strlen(s), p=1;
for(int i=0;i<len;i++)
{
int ch=s[i]-'a';
if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
p=trie[p].nxt[ch];
}
return (trie[p].ed)?(trie[p].ed):(trie[p].ed=++idcnt);
}
};
int par[2*maxn],ran[2*maxn];
int find(int x){return (par[x]==x)?x:(par[x]=find(par[x]));}
void unite(int x,int y)
{
x=find(x), y=find(y);
if(x==y) return;
if(ran[x]<ran[y]) par[x]=y;
else par[y]=x, ran[x]+=(ran[x]==ran[y]);
}
int degree[2*maxn];
int main()
{
Trie::init();
memset(degree,0,sizeof(degree));
for(int i=1;i<2*maxn;i++) par[i]=i,ran[i]=0;
char s[2][13];
while(scanf("%s %s",s[0],s[1])!=EOF)
{
int u=Trie::insert(s[0]);
int v=Trie::insert(s[1]);
if(find(u)!=find(v)) unite(u,v);
degree[u]++;
degree[v]++;
}
int cnt=0;
int z=find(1);
for(int i=1;i<=Trie::idcnt;i++)
{
if(find(i)!=z) {
printf("Impossible\n");
return 0;
}
if(degree[i]%2) cnt++;
if(cnt>2) {
printf("Impossible\n");
return 0;
}
}
if(cnt==1) {
printf("Impossible\n");
return 0;
} else {
printf("Possible\n");
return 0;
}
}
