ACM: trie树 poj 2513

Colored Sticks

Description

You are given a bunch of woodensticks. Each endpoint of each stick is colored with some color. Isit possible to align the sticks in a straight line such that thecolors 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 thecolors of the endpoints of one stick. A word is a sequence oflowercase letters no longer than 10 characters. There is no morethan 250000 sticks.

Output

If the sticks can be aligned inthe desired way, output a single line saying Possible, otherwiseoutput Impossible.

Sample Input

blue red
red violet
cyan blue
blue magenta
magenta cyan

Sample Output

Possible

题意: 给你一捆木头, 木头两端是有颜色, 现在要求将所有的木头首尾相接, 但是相接的端口要去
         颜色一样.

解题思路:
         1. trie树: 利用串的公共前缀来节约内存, 加快检索速度.
         性质:
                   1). 根节点不包含字母, 除了根节点外每一个节点都仅包含一个字符.
                   2). 从根到某一节点, 路径上结果的字母依次连起来就是一个字母序列,就是一个串.
                   3). 每个节点的所有子节点的字母都不相同.
          实现方法:
                   每个节点设置三个域: id, flag, next[26]
                   id域用来保存字母, flag域来标记到当前节点是否形成一个串, next[26]下一个节点指针域.
          如图:
          2. 题目要求木头首尾相接, 可以联想到blue-red -> red-cyan -> cyan-magenta -> magenta-blue
              很像一条路径, 欧拉回路.无向欧拉: 每个节点的度%2 == 0或者有且仅有2个点的度数为奇数.
              当然欧拉路采用是并查集的压缩路径
          3. 欧拉路的判断 + trie树的结构查找题目就解决了.

代码:
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 500005

struct node
{
    int id;
    bool flag;
    node *next[27];
}root;

int d[MAX], p[MAX];
int num;
char str1[11], str2[11];

void init()
{
    num = 0;
    for(int i = 0; i < MAX; ++i)
    {
        d[i] = 0;
        p[i] = i;
    }
    root.flag = false;
    root.id = 0;
    memset(root.next, 0, sizeof(root.next));
}

int find(int x)
{
    return p[x] == x ? x : (p[x] = find(p[x]));
}

void union_set(int x, int y)
{
    x = find(x);
    y = find(y);
    if(x != y) p[y] = x;
}

int insert(char *str)
{
    node *p = &root;

    int len = strlen(str);
    int temp;
    for(int i = 0; i < len; ++i)
    {
        temp = str[i] - 'a';
        if( !p->next[temp] )
        {
            p->next[temp] = new node;
            p->next[temp]->flag = false;
            p->next[temp]->id = 0;
            memset(p->next[temp]->next, 0, sizeof(p->next[temp]->next));
        }
        p = p->next[temp];
    }
    
    if(p->flag) return p->id;
    else
    {
        p->flag = true;
        p->id = ++num;
        return p->id;
    }
}

int main()
{
//    freopen("input.txt","r",stdin);
    init();
    while( scanf("%s %s",str1, str2) != EOF)
    {
        int x = insert(str1);
        int y = insert(str2);
        
        d[x]++;
        d[y]++;
        union_set(x, y);
    }
    
    int head = find(1);
    int t = 0;
    bool flag = false;
    for(int i = 1; i <= num; ++i)
    {
        if(d[i] % 2 != 0)
            t++;
        if(t > 2)
        {
            printf("Impossible\n");
            flag = true;
            break;
        }
        
        if(find(i) != head)
        {
            printf("Impossible\n");
            flag = true;
            break;
        }
    }
    
    if( !flag )
    {
        if(t == 1)
            printf("Impossible\n");
        else
            printf("Possible\n");
    }    
    return 0;
}