POJ2513 并查集+欧拉路径+trie树

Posted 2022-06-08 yuanweidao

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 

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