poj 2513 Colored Sticks trie树+欧拉图+并查集

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Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 27955		Accepted: 7403

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

Hint

Huge input,scanf is recommended.

一开始想用map建立字符串和数字的映射，然后用并查集判断联通，欧拉图判断是否可以构成一笔画问题，但是最后超时了，网上一查才知道必须用trie建立映射关系才能过

因为字符最多10位，那么树的深度最多10层，映射任何一个字符串都可以在近似常数时间内完成，效率非常高

先附上map实现的TLE的代码：

#include<stdio.h>
#include<string>
#include<map>
using namespace std;
int father[500010];
int degree[500010] = {0};
//int *root[26]; //
int getfather(int i)
{
	if(father[i] == 0)
		return father[i] = i;
	else if(father[i] == i)
		return i;
	else
		return father[i] = getfather(father[i]);
}
void mergeset(int a, int b)
{
	a = getfather(a);
	b = getfather(b);
	father[a] = b;
}
int main()
{
//	freopen("in.txt", "r", stdin);
	map<string, int> color;

	char color1[20], color2[20];
	int total = 0;
	while(scanf("%s%s", color1, color2) != EOF)
	{
		int a, b;
		if(color[color1] == 0)
			a = (color[color1] = ++total);
		if(color[color2] == 0)
			b = (color[color2] = ++total);
		degree[a] ++;
		degree[b] ++;
		mergeset(a, b);
	}
	int i;
	int fa = getfather(1);
	bool flag = 0;
	int count = 0;
//	for(i = 1; i <
	for(i = 1; i <= total; i++)
	{
		if(getfather(i) != fa)
			break;
		if(degree[i] % 2 != 0)
		{
			count ++;
			if(count > 2)
				break;
		}
	}
	if(i <= total || count == 1)
		printf("Impossible\n");
	else
		printf("Possible\n");

	return 0;
}

可以发现map省事很多，但是效率不高

这是trie图的AC代码：1250MS

#include<stdio.h>
#include<string>
#include<iostream>
#include<map>
using namespace std;
int father[500010];
int degree[500010] = {0};

struct Node
{
	int num;
	Node *next[26];
	Node()
	{
		num = 0;
		int i;
		for(i = 0; i < 26; i++)
			next[i] = 0;
	}
};
Node root[26];
int total;
int getnum(char * str, Node * node)//getnum函数既可以插入也可以获取，如果没有这个单词就建立一个，如果有了就返回这个单词的编号
{
	if(*str == 0)
	{
		if(node->num != 0)
			return node->num;
		else
			return node->num  = ++total;
	}
	if(node->next[*str - 'a'] == 0)
		node->next[*str - 'a'] = new Node;
	return getnum(str + 1, node->next[*str - 'a']);
}
int getfather(int i)
{
	if(father[i] == 0)
		return father[i] = i;
	else if(father[i] == i)
		return i;
	else
		return father[i] = getfather(father[i]);
}
void mergeset(int a, int b)
{
	a = getfather(a);
	b = getfather(b);
	father[a] = b;
}
int main()
{
//	freopen("in.txt", "r", stdin);
	char color1[20], color2[20];
	while(scanf("%s%s", color1, color2) != EOF)
	{
		int a = getnum(color1 + 1, &root[*color1 - 'a']), b = getnum(color2 + 1, &root[*color2 - 'a']);
		degree[a] ++;
		degree[b] ++;
		mergeset(a, b);
	}
	int i;
	int fa = getfather(1);
	bool flag = 0;
	int count = 0;
	for(i = 1; i <= total; i++)
	{
		if(getfather(i) != fa)
			break;
		if(degree[i] % 2 != 0)
		{
			count ++;
			if(count > 2)
				break;
		}
	}
	if(i <= total || count == 1)
		printf("Impossible\n");
	else
		printf("Possible\n");

	return 0;
}