POJ 2513 Colored Sticks （Trie + 并查集 + 欧拉通路）

题目大意：http://poj.org/problem?id=2513

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 redred violetcyan blueblue magentamagenta cyan

Sample Output

Possible

解题思路：可以考虑把每个颜色建成一个节点，题目要求把所有的木棍拼成一条直线，所以也就是说除了两端的端点，中间的节点的度数都是偶数，所以题目也就变成了能否构成这么一条欧拉通路，因为题目数据比较大，所以不能使用map，只能使用字典树，统计每个节点的度数。不过最后需要判断一下是否是连通图，所以需要用并查集，判断一下。

AC代码：

/*    @Author: wchhlbt    @Date:   2017/4/18*///#include <bits/stdc++.h>#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>#include <limits>#include <climits>#include <cstdio>#define Fori(x) for(int i=0;i<x;i++)#define Forj(x) for(int j=0;j<x;j++)#define maxn 520000#define inf 0x3f3f3f3f#define ONES(x) __builtin_popcount(x)using namespace std;typedef long long ll ;const double eps =1e-8;const int mod = 1000000007;typedef pair<int, int> P;const double PI = acos(-1.0);int dx[4] = {0,0,1,-1};int dy[4] = {1,-1,0,0};int color,cnt;int deg[maxn];//数组式写法的字典树struct Node{   int num;//记录每个单词出现的次数   int pNext[26];}node[maxn<<4];int Insert(char s[]){    int p = 0;    int len = strlen(s);    for(int i = 0; i<len; i++){       int m = s[i] - 'a';        if (!node[p].pNext[m]) {            node[p].pNext[m] = ++cnt;            for(int j = 0; j<26; j++)                node[cnt].pNext[j] = 0;            node[cnt].num = 0;       }       p = node[p].pNext[m];   }   if(node[p].num==0)        node[p].num = color++;   return node[p].num;}//并查集int city[maxn];int getcity(int point){    if(city[point] == point)    {        return point;    }    else    {        city[point] = getcity(city[point]);        return city[point];    }}bool connect(int point1,int point2){    int t1,t2;    t1 = getcity(point1);    t2 = getcity(point2);    if(t1!=t2)    {        city[t2] = t1;        return true;    }    return false;}void init(){    for(int i = 1; i<=maxn-2; i++)    {        city[i] = i;    }}int main(){    color = 1;    init();    char str[20];    while(~scanf("%s",str))    {        int id1 = Insert(str);        scanf("%s",str);        int id2 = Insert(str);        deg[id1]++; deg[id2]++;        connect(id1,id2);    }    int k = 0;    for(int i = 1; i<color; i++)    {        if(deg[i]%2)            k++;    }    if(k!=0 && k!=2){        printf("Impossible\n");    }    else{        int ans = getcity(1);        for(int i = 1; i<color; i++){            if(getcity(i)!=ans){                printf("Impossible\n");                return 0;            }        }        printf("Possible\n");    }    return 0;}