Colored Sticks

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Description

Input

Output

Sample Input

blue red
red violet
cyan blue
blue magenta
magenta cyan

Sample Output

Possible

题解：就是判断欧拉回路。这是无向图。一个欧拉回路每条边经过一次，那么依次把边拉直，刚好就是一根直线（半欧拉回路）。判断欧拉回路需要在连通图的条件下判断度数是基数的个数。为0是欧拉回路，2是半欧拉回路。判断是否联通肯定就是用并查集了。此题用map超时。所以需要用字典树为每个字符串一个编号。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string> 
#include <map>

using namespace std;

struct Node
{
	int num;
	Node* next[26];
	Node()
	{
		num = 0;
	    for(int i = 0;i < 26;i++)
		{
			next[i] = NULL;
		 } 
	}
};

Node* root;
int degree[3000006];
int pre[3000006];
int k = 1;

int insert(char* s)
{
	int len = strlen(s);
	Node* p = root;
	for(int i = 0;i < len;i++)
	{
		int x = s[i] - 'a';
		if(p->next[x] == NULL)
		{
			Node* q = new Node();
			p->next[x] = q;
		}
		p = p->next[x];
	}
	if(p->num == 0)
	{
		p->num = k++;
	}
	degree[p->num]++;
	return p->num;
}

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

int main()
{
	char s1[100];
	char s2[100];
	root = new Node();
	memset(degree,0,sizeof(degree));
	for(int i = 1;i <= 250000;i++)
	{
		pre[i] = i;
	}
	while(scanf("%s%s",s1,s2) != EOF)
	{
		int x = insert(s1);
		int y = insert(s2);
		x = find(x);
		y = find(y);
		pre[x] = y;
	}
	
	int x = find(1);
	for(int i = 1;i < k;i++)
	{
		if(x != find(pre[i]))
		{
			printf("Impossible\n");
			return 0;
		}
	}
	int cnt = 0;
	for(int i = 1;i < k;i++)
	{
		//cout<<degree[i]<<endl;
		if((degree[i] & 1) != 0)
		{
			cnt++;
		}
	}
	if(cnt == 0 || cnt == 2)
	{
		printf("Possible\n");
	}
	else
	{
		printf("Impossible\n");
	}

	return 0;
}

Colored Sticks

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原文地址：http://blog.csdn.net/wang2534499/article/details/48053375