poj 2513 -- Colored Sticks

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 30037		Accepted: 7924

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.

 

 

   

     题目大意：给定一捆木棒，每根木棒的每个端点涂有某种颜色。问：是否能将这些棍子首尾相连，排成一条直线，且相邻两根棍子的连接处端点的颜色一样。

 

     思路：首先用字典树（Trie）将每个单词编号。用并查集判断是否能连通。然后求解欧拉通路。

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　　定理：无向图G存在欧拉通路的充要条件是：G为连通图，并且G仅有两个奇度结点或者无奇度结点。

　　推论1：当G是仅有两个奇度结点的连通图时，G的欧拉通路必以此两个结点为端点。

　　推论2：当G是无奇度结点的连通图时，G必有欧拉回路。

　　推论3：G为欧拉图的充分必要条件是G为无奇度结点的连通图。

 

　　定理：有向图D存在欧拉通路的充要条件是：D为有向图，D的基图连通，并且所有顶点的出度与入度都相等；或者除两个顶点外，其余顶点的出度与入度都相等，而这两个顶点中一个顶点的出度与入度之差为1,另一个顶点的出度与入度之差为-1.

　　推论1：当D除 出入度之差为1,-1的两个顶点之外，其余顶点的出度与入度都相等时，D的有向欧拉通路必以出入度之差为1的顶点作为始点，以出入度之差为-1的顶点作为终点。

　　推论2：当D的所有顶点的出入度都相等时，D中存在有向欧拉回路。

　　推论3：有向图D为有向欧拉图的充分必要条件是D的基图为连通图，并且所有顶点的出入度都相等。

　　　

代码如下：

 

  1 /*======================================================================
  2  *           Author :   kevin
  3  *         Filename :   ColoredSticks.cpp
  4  *       Creat time :   2014-07-31 10:13
  5  *      Description :
  6 ========================================================================*/
  7 #include <iostream>
  8 #include <algorithm>
  9 #include <cstdio>
 10 #include <cstring>
 11 #include <queue>
 12 #include <cmath>
 13 #define clr(a,b) memset(a,b,sizeof(a))
 14 #define M 500005
 15 using namespace std;
 16 struct Trie{
 17     Trie *next[26];
 18     int num;
 19 };
 20 Trie *root;
 21 int father[M+5],d[M+5];
 22 int cnt = 1;
 23 int CreateTrie(char *str)
 24 {
 25     Trie *p = root,*q;
 26     for(int i = 0; str[i]; i++){
 27         int id = str[i] - 'a';
 28         if(p->next[id] == NULL){
 29             q = (Trie *)malloc(sizeof(Trie));
 30             for(int j = 0; j < 26; j++){
 31                 q->next[j] = NULL;
 32                 q->num = 0;
 33             }
 34             p->next[id] = q;
 35             p = p->next[id];
 36         }
 37         else{
 38             p = p->next[id];
 39         }
 40     }
 41     if(!(p->num))
 42         p->num = cnt++;
 43     return p->num;
 44 }
 45 
 46 void DelTrie(Trie *T)
 47 {
 48     int i;
 49     if(T == NULL) return;
 50     for(i = 0; i < 26; i++){
 51         if(T->next[i] != NULL){
 52             DelTrie(T->next[i]);
 53         }
 54     }
 55     free(T);
 56 }
 57 int FindSet(int x)
 58 {
 59     if(x != father[x]){
 60         father[x] = FindSet(father[x]);
 61     }
 62     return father[x];
 63 }
 64 void Union(int x,int y)
 65 {
 66     x = FindSet(x);
 67     y = FindSet(y);
 68     father[x] = y;
 69 }
 70 int main(int argc,char *argv[])
 71 {
 72     char str1[15],str2[15];
 73     for(int i = 0; i < M; i++){
 74         father[i] = i;
 75         d[i] = 0;
 76     }
 77     root = (Trie*)malloc(sizeof(Trie));
 78     for(int i = 0; i < 26; i++){
 79         root->next[i] = NULL;
 80         root->num = 0;
 81     }
 82     while(scanf("%s %s",str1,str2)!=EOF){
 83         int x1 = CreateTrie(str1);
 84         int x2 = CreateTrie(str2);
 85         Union(x1,x2);
 86         d[x1]++;
 87         d[x2]++;
 88     }
 89     int cnt_father = 0,cnt_d = 0;
 90     int t = FindSet(1);
 91     for(int i = 1; i < cnt; i++){
 92         int x = FindSet(i);
 93         if(x == t) cnt_father++;
 94         if(d[i] % 2){
 95             cnt_d++;
 96         }
 97     }
 98     if(cnt_father != cnt-1 || !(cnt_d == 0 || cnt_d == 2)){
 99         printf("Impossible\n");
100     }
101     else{
102         printf("Possible\n");
103     }
104     DelTrie(root);
105     return 0;
106 }