POJ2513 Colored Sticks(Trie+欧拉回路)

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题意:给你n(n<=250000)根小木棒,木棒两端染有一些颜色,两根木棒相连需要保证相连端点颜色一致.求这些木棒是否可以全部相连?题解:可以将木棒理解为边,连接两种颜色,那么全部相连就变成了一笔画的问题,即判断欧拉回路.欧拉回路需要满足两个条件:1.度数为奇数的点为0个或2个2.图联通现在的问题是如何把字符串变成数字的点,用map是肯定会TLE的hash应该可以,但tire好写,所以我选择了trie树,存入trie的值为该字符串的映射图联通用并查集去判断.代码如下:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int cnt=0,du[500010],fa[500050];

struct Trie
{
    int sz,tr[500010][26];
    int time[500010];

    void clear()
    {
        sz=0;
        memset(tr,0,sizeof(tr));
    }

    void insert(char* c,int x)
    {
        int rt=0,len=strlen(c);
        for(int i=0;i<len;i++)
        {
            if(!tr[rt][c[i]-‘a‘])
            {
                tr[rt][c[i]-‘a‘]=++sz;
            }
            rt=tr[rt][c[i]-‘a‘];
        }
        time[rt]=x;
    }

    int search_t(char *c)
    {
        int rt=0,len=strlen(c);
        for(int i=0;i<len;i++)
        {
            if(!tr[rt][c[i]-‘a‘])
            {
                return 0;
            }
            rt=tr[rt][c[i]-‘a‘];
        }
        return time[rt];
    }

}trie;

void mem(int n)
{
    for(int i=1;i<=n;i++)
    {
        fa[i]=i;
    }
}

int find(int x)
{
    if(x!=fa[x])
    {
        fa[x]=find(fa[x]);
    }
    return fa[x];
}

void union_(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    {
        fa[fx]=fy;
    }
}

int euler()
{
    int sum=0,t=find(1);
    for(int i=1;i<=cnt;i++)
    {
        if(du[i]%2==1)
        {
            sum++;
        }
    }
    if(sum!=0&&sum!=2)
    {
        return 0;
    }
    for(int i=1;i<=cnt;i++)
    {
        if(find(i)!=find(t))
        {
            return 0;
        }
    }
    return 1;
}

int main()
{
    char s1[20],s2[20];
    mem(500000);
//    freopen("1.in","r",stdin);
//    freopen("1.out","w",stdout);
    while(scanf("%s %s\n",s1,s2)!=EOF)
    {
        int from,to;
        if(!trie.search_t(s1))
        {
            trie.insert(s1,++cnt);
        }
        from=trie.search_t(s1);
        du[from]++;
        if(!trie.search_t(s2))
        {
            trie.insert(s2,++cnt);
        }
        to=trie.search_t(s2);
        du[to]++;
        union_(from,to);
    }
    if(euler())
    {
        puts("Possible");
    }
    else
    {
        puts("Impossible");
    }
}

原文地址：https://www.cnblogs.com/stxy-ferryman/p/8496188.html