POJ2513 【并查集+欧拉路径+trie树】

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个点的度为奇

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。

 #include<stdio.h>
 #include<string.h>
 #define mem(a, b) memset(a, b, sizeof(a))
 const int MAXN =  *  + ;//最多MAXN种颜色 即最多MAXN种不同的点 

 char s1[], s2[];
 int deg[MAXN];
 int trie[MAXN][], cnt, tot;
 int id[MAXN], pre[MAXN];

 int insert(char s[])
 {
     int flag = ;
     int len = strlen(s);
     int root = ;
     for(int i = ; i < len; i ++)
     {
         int id = s[i] - 'a';
         if(!trie[root][id])
         {
             flag = ;  //单词之前没出现过
             trie[root][id] = ++ cnt;
         }
         root = trie[root][id];
     }
     if(!flag)
     {
         tot ++;
         id[root] = tot;
         return id[root];
     }
     else
         return id[root];
 }

 int find(int x)
 {
     if(pre[x] == x)
         return x;
     else
     {
         int root = find(pre[x]);
         pre[x] = root;
         return pre[x];
     }
 }

 int main()
 {
     for(int i = ; i <= MAXN + ; i ++)
         pre[i] = i;
     while(scanf("%s%s", s1, s2) != EOF)
     {
         int a = insert(s1), b = insert(s2);//返回的是颜色所对应的序号
         deg[a] ++, deg[b] ++; //点的度数 判断是否存在欧拉路径
         int x = find(a), y = find(b); //查找根节点
         if(x != y)
             pre[y] = x;
     }
     int flag = ;
     for(int i = ; i < tot; i ++) //总共有tot个不同的点
         if(find(i) != find(i + ))
         {
             flag = ;
             break;
         }
     if(flag == )
         printf("Impossible\n");
     else
     {
         int xx = ;
         for(int i = ; i <= tot; i ++)
             if(deg[i] % )
                 xx ++;
         if(xx ==  || xx == )
             printf("Possible\n");
         else
             printf("Impossible\n");
     }
     return ;
 }