POJ2513【并查集+字典树+欧拉】

/*Colored Sticks 
Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 256000/128000K (Java/Other)
Total Submission(s) : 0   Accepted Submission(s) : 0
Problem 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 

Source
PKU
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
typedef struct Tree
{
	bool flag;
	int id;
	Tree * next[27];
}Tree;
Tree root;
int colorid, colornum, f[500010], color[500010];
int getid(char *s)
{
	Tree *p = &root, *q;
	int i, j, k;
	k = strlen(s);
	for(i = 0; i < k; ++i)
	{
		j = s[i] - 'a';//id
		if(p->next[j] == NULL)
		{
			q = (Tree *)malloc(sizeof(root));
			q->flag = false;
			for(int l = 0; l < 26; ++l)
			q->next[l] = NULL;
			p->next[j] = q;
		}
		p = p->next[j];
		if(i == k-1)
		break;
	}
	if(!p->flag)
	{
		p->flag = true;
		p->id = ++colorid;
	}
	return p->id;
}
int find(int x)
{
	if(x != f[x])
	f[x] = find(f[x]);
	return f[x];
}
bool merge(int x, int y)
{
	x = find(x);
	y = find(y);
	if(x != y)
	{
		f[x] = y;
		return true;
	}
	return false;
}
int main()
{
	char s1[11], s2[11];
	int i, j, k;
	for(i = 1; i <= 500001; ++i)
	f[i] = i;
	colorid = 0;
	colornum = 0;
	memset(color, 0, sizeof(color));
	for(i = 0; i < 26; ++i)//字典树根节点初始化 
		root.next[i] = NULL;
	while(scanf("%s%s", s1, s2) != EOF)
	{
		i = getid(s1);
		j = getid(s2);
		color[i]++;//保存度数 
		color[j]++;
		if(merge(i, j))//如果合并成功 
		colornum++;
	}
	if(colorid == 0)
	{
		printf("Possible\n");
		return 0;
	}
	if(colornum != colorid-1)//如果不连通 
	{
		printf("Impossible\n");
	}
	else//连通 
	{
		j = 0;
		for(i = 1; i <= colorid; ++i)
		{
			if(color[i] & 1)//记录奇数度数个数 
			j++;
		}
		if(j == 1 || j >= 3)
		printf("Impossible\n");
		else
		printf("Possible\n");
	}
	return 0;
}

题意：给出一个筷子，筷子的两端都涂有一种颜色，现在给出n根两端都涂有颜色的筷子，当2根筷子的一端具有同种颜色时，两根筷子可以连成一条直线，问这n根筷子是否能连成一条直线。

思路：本题可以将颜色看做是一个节点，筷子看做是边，则题目的要求就类似于，给出n个节点，和m条边，问是否能一笔画走完每一条边一次。这就是欧拉回路，所以根据定义，若是所有的颜色的边的个数全为偶数，或是只有2个颜色的边数为奇数，则可以一笔画走完。现在就要开始模拟这种思想。但是这里还有一个难点就是怎么给颜色字符串编号的问题，如果按照常规的方法必定超时，所以这里又要用到字典树，对颜色进行编号，可以大大提高效率。对每种颜色进行编号的同时可以进行边数的统计，以及并查集的操作，最后用来判断这n种颜色是否能够连通，并且边数是否符合要求，最终判定是否是欧拉回路。

体会：第一做这种综合性的题目，难度好大0.0