题目链接：http://poj.org/problem?id=2513
Colored Sticks
| Time Limit: 5000MS | Memory Limit: 128000K | |
| Total Submissions:40949 | Accepted: 10611 |
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.
题目大意：给出以下字符串，代表每根棍子的两端颜色，只能将颜色相同的端点连接起来。问是否能连成一欧拉路径。
思路：
1.棍子两端无方向要求，故是无向图的欧拉路径，无向图判断欧拉路径的充要条件是：图连通（无向图的连通性可以用并查集来判断，但有向图不可以）+ 点的度全为偶/当且仅当只有2个
2.关键在于将颜色字符串映射成整型编号，有两种思路：用map直接映射（TLE），用trie树给字符串上编号。
3.具体实现在代码中，值得注意的一点是 G++才能AC。
1 #include<stdio.h>
2 #include<string.h>
3 #define mem(a, b) memset(a, b, sizeof(a))
4 const int MAXN = 250000 * 2 + 10;//最多MAXN种颜色 即最多MAXN种不同的点
5
6 char s1[12], s2[12];
7 int deg[MAXN];
8 int trie[MAXN][27], cnt, tot;
9 int id[MAXN], pre[MAXN];
10
11 int insert(char s[12])
12 {
13 int flag = 1;
14 int len = strlen(s);
15 int root = 0;
16 for(int i = 0; i < len; i ++)
17 {
18 int id = s[i] - ‘a‘;
19 if(!trie[root][id])
20 {
21 flag = 0; //单词之前没出现过
22 trie[root][id] = ++ cnt;
23 }
24 root = trie[root][id];
25 }
26 if(!flag)
27 {
28 tot ++;
29 id[root] = tot;
30 return id[root];
31 }
32 else
33 return id[root];
34 }
35
36 int find(int x)
37 {
38 if(pre[x] == x)
39 return x;
40 else
41 {
42 int root = find(pre[x]);
43 pre[x] = root;
44 return pre[x];
45 }
46 }
47
48 int main()
49 {
50 for(int i = 1; i <= MAXN + 1; i ++)
51 pre[i] = i;
52 while(scanf("%s%s", s1, s2) != EOF)
53 {
54 int a = insert(s1), b = insert(s2);//返回的是颜色所对应的序号
55 deg[a] ++, deg[b] ++; //点的度数 判断是否存在欧拉路径
56 int x = find(a), y = find(b); //查找根节点
57 if(x != y)
58 pre[y] = x;
59 }
60 int flag = 1;
61 for(int i = 1; i < tot; i ++) //总共有tot个不同的点
62 if(find(i) != find(i + 1))
63 {
64 flag = 0;
65 break;
66 }
67 if(flag == 0)
68 printf("Impossible\n");
69 else
70 {
71 int xx = 0;
72 for(int i = 1; i <= tot; i ++)
73 if(deg[i] % 2)
74 xx ++;
75 if(xx == 0 || xx == 2)
76 printf("Possible\n");
77 else
78 printf("Impossible\n");
79 }
80 return 0;
81 }
原文地址：https://www.cnblogs.com/yuanweidao/p/11296325.html