POJ2513Colored Sticks（欧拉通路）（字典树)(并查集）

                                                         Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 35612		Accepted: 9324

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
【分析】欧拉通路的判定。一开始用map给字符串编号，超时了，后来在网上烤来一份字典树hash.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=5000005;
const int M=1005;
ll power(ll a,int b,ll c){ll ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;}
int parent[N];
const int MAX=26;
int degree[N];//度数
int color;//颜色编号，从0开始，最后就是颜色总数
int Find(int x) {
    if(parent[x] != x) parent[x] = Find(parent[x]);
    return parent[x];
}//查找并返回节点x所属集合的根节点
void Union(int x,int y) {
    x = Find(x);
    y = Find(y);
    if(x == y) return;
    parent[y] = x;
}//将两个不同集合的元素进行合并
struct Trie
{
    bool isWord;
    struct Trie *next[MAX];
    int id;
};
int insert(Trie *root,char *word)//返回颜色编号
{
    Trie *p=root;
    int i=0;
    while(word[i]!='')
    {
        if(p->next[word[i]-'a']==NULL)
        {
            Trie *temp=new Trie;
            temp->isWord=false;
            for(int j=0;j<MAX;j++)
              temp->next[j]=NULL;
            temp->isWord=false;
            temp->id=0;
            p->next[word[i]-'a']=temp;
        }
        p=p->next[word[i]-'a'];
        i++;
    }
    if(p->isWord)
    {
        return p->id;
    }
    else
    {
        p->isWord=true;
        p->id=color++;
        return p->id;
    }
}
void del(Trie *root)
{
    for(int i=0;i<MAX;i++)
    {
        if(root->next[i]!=NULL)
         del(root->next[i]);
    }
    free(root);
}
int main()
{
    char str1[20],str2[20];
    Trie *root=new Trie;
    for(int i=0;i<MAX;i++)
      root->next[i]=NULL;
    root->isWord=false;
    root->id=0;//初始化
    color=0;
    for(int i=0;i<N;i++)parent[i]=i;
    memset(degree,0,sizeof(degree));
    while(scanf("%s%s",&str1,&str2)!=EOF)
    {
        int t1=insert(root,str1);
        int t2=insert(root,str2);
     //   printf("%d %d
",t1,t2);
        degree[t1]++;
        degree[t2]++;
        Union(t1,t2);
    }
    int cnt1=0,cnt2=0;
    for(int i=0;i<color;i++)
    {
        if(parent[i]==i)cnt1++;
        if(degree[i]%2==1)cnt2++;
        if(cnt1>1)break;
        if(cnt2>2)break;
    }
    //数据比较坑人，存在0根木棒的情况，此时cnt1==0
    if((cnt1==0||cnt1==1)&&(cnt2==0||cnt2==2))
      printf("Possible
");
    else printf("Impossible
");
    //del(root);//单组输入可以不释放空间，可以节省时间
    return 0;

}