poj2531(trie + 欧拉路径判定)

思路：这问题显然是要判断欧拉路径的存在，然后要把串看成一个点，然后统计串的种类（trie树），最后就是并查集来判断连通性 ＋ 度的判断（无向图的欧拉路径 ＝＝ 度为奇数的点的个数要么为0，要么为2，不然不存在欧拉路径，这个是一个充要条件）

点击题目链接

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define MEM(a,b) memset(a,b,sizeof a)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct node{
int val;
node* nxt[26];
}*rt,memery[5100050];
int F[510005],deg[510005],cnt,ant;
node *newnode(){
node *p = &memery[ant++];
for (int i = 0;i < 26;i++){
p->nxt[i] = NULL;
}
return p;
}
void insert(char *s,int x){
node* p = rt;
for (int i = 0;s[i];i++){
int k = s[i] - 'a';
if (p->nxt[k] == NULL){
p->nxt[k] = newnode();
}
p = p->nxt[k];
}
p->val = x;
}
int search(char *s){
node *p = rt;
for (int i = 0;s[i];i++){
int k = s[i] - 'a';
if (p->nxt[k] == NULL) return 0;
p = p->nxt[k];
}
return p->val;
}
void init(){
for (int i = 1;i <= 510000;i++)
F[i] = i;
MEM(deg, 0);
cnt = 0;
}
int Find(int x){
if (x == F[x]) return x;
else return F[x] = Find(F[x]);
}
void Union(int x,int y){
F[x] = y;
}
bool Judge(){
int odd = 0;
for (int i = 1;i <= cnt;i++){
if (deg[i] & 1) odd++;
}
if (odd != 0 && odd != 2) return false;
int k = Find(1);
for (int i = 2;i <= cnt;i++){
if (k != Find(i)) return false;
}
return true;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
init();
rt = newnode();
char s1[15],s2[15];
while(scanf("%s %s",s1,s2) != EOF){
int x = search(s1);
int y = search(s2);
if (x == 0) insert(s1,x = ++cnt);
if (y == 0) insert(s2,y = ++cnt);
deg[x]++;
deg[y]++;
int t1 = Find(x);
int t2 = Find(y);
if (t1 != t2) Union(t1,t2);
}
if (Judge()) printf("Possible\n");
else printf("Impossible\n");
return 0;
}