Colored Sticks （并查集+Trie + 欧拉路）

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 37340		Accepted: 9796

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.

Source

The UofA Local 2000.10.14

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 500008
#define INF 1000000009
#define eps 0.00000001
/*
欧拉路 一笔画问题！
Trie树就是一种节省空间的hash
利用并查集判断是否联通
*/
char a[15], b[15];
int cnt = 0, index = 1, pre[MAXN], degree[MAXN];
typedef struct node
{
    bool flag;
    int id;
    struct node* next[26];
}*Tree;
node memory[MAXN];
Tree Newnode()
{
    Tree T = &memory[cnt++];
    T->flag = false;
    T->id = -1;
    for (int i = 0; i < 26; i++)
        T->next[i] = NULL;
    return T;
}
int Insert(char* s, Tree T)
{
    int p = 0;
    while (s[p] != '\0')
    {
        int k = s[p] - 'a';
        if (!T->next[k])
            T->next[k] = Newnode();
        T = T->next[k];
        p++;
    }
    if (T->flag)
        return T->id;
    else
    {
        T->flag = true;
        T->id = index++;
        return T->id;
    }
}
int find(int x)
{
    if (pre[x] == -1)
        return x;
    else
        return pre[x] = find(pre[x]);
}
void mix(int a, int b)
{
    int fa = find(a), fb = find(b);
    if (fa != fb)
    {
        pre[fa] = fb;
    }
}
int main()
{
    memset(pre, -1, sizeof(pre));
    memset(degree, 0, sizeof(degree));
    Tree T = Newnode();
    while (scanf("%s %s", a, b) != EOF)
    {
        //if (a[0] == '#') break;
        int ia = Insert(a, T), ib = Insert(b, T);
        //cout << ia << ' ' << ib << endl;
        degree[ia]++;
        degree[ib]++;
        mix(ia, ib);
    }
    int s = find(1), tmp = 0;
    for (int i = 1; i < index; i++)
    {
        if (find(i) != s)   //存在多个祖先，图为森林，不连通  
        {
            printf("Impossible\n");
            return 0;
        }
        if (degree[i] % 2)
            tmp++;
    }
    if(tmp==0 || tmp==2)
        printf("Possible\n");
    else
        printf("Impossible\n");
}