poj 2513 Colored Sticks (trie树+并查集+欧拉路）

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 40043		Accepted: 10406

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.

 

题目大意：

给定n根木棒，首尾各有一种颜色。求问有没有可能，将这些木棒连成一排，使得接头处颜色相同。

 

判断无向图是否存在欧拉路经典题。

一根木棒其实就是给两种颜色连上一条无向边（可能会有重边的）。那么一条欧拉路就相当于一种连接方式。

无向图存在欧拉图的条件：

1）联通。这个交给并查集就好了。

2）度数为奇数的点只能是0个或2个。这个建图的时候或建完图后都挺好处理的。

主要是如何建图呢。字符串太多了，要hash成数。用字典树比较合适。

每次遇见一种颜色，看字典树中有没有该颜色。如果有，返回该颜色的id，否则加入该颜色，id取全局变量++col。

 

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>
#include<deque>
typedef long long ll;
const int maxn=250000;

int col;
struct ttrie
{
    bool isword;
    int id;
    ttrie* next[26];
    ttrie()
    {
        isword=false;
        for(int i=0;i<26;i++)
            next[i]=NULL;
    }
};

char str1[15],str2[15];

int add(ttrie* root,char str[])
{
    ttrie *p=root,*q;
    for(int i=0;str[i]!='';i++)
    {
        int temp=str[i]-'a';
        if(p->next[temp]==NULL)
        {
            q=new ttrie;
            p->next[temp]=q;
            p=q;
        }
        else
            p=p->next[temp];
    }
    if(p->isword)
        return p->id;
    else
    {
        p->isword=true;
        p->id=++col;
        return p->id;
    }
}

int degree[maxn*2+5];
int cnt;
struct tedge
{
    int u,v;
};
tedge edge[maxn+5];

int fa[maxn*2+5];

int getfa(int x)
{
    if(fa[x]==x)
        return x;
    else
        return fa[x]=getfa(fa[x]);
}

int main()
{
    col=0;
    ttrie* root=new ttrie;

    memset(degree,0,sizeof(degree));
    cnt=0;
    while(scanf("%s%s",str1,str2)!=EOF)
    {
        int u=add(root,str1);
        int v=add(root,str2);
        degree[u]++;
        degree[v]++;
        edge[++cnt]=(tedge){u,v};
    }

    bool flag=true;
    for(int i=1;i<=col;i++)
        fa[i]=i;
    int tot=col;
    for(int i=1;i<=cnt;i++)
    {
        int fx=getfa(edge[i].u);
        int fy=getfa(edge[i].v);
        if(fx!=fy)
        {
            fa[fx]=fy;
            tot--;
        }
    }
    if(tot>1)
        flag=false;
    if(flag)
    {
        int odd=0;
        for(int i=1;i<=col;i++)
        {
            if(degree[i]%2)
                odd++;
        }
        if(odd!=0&&odd!=2)
            flag=false;
    }
    if(flag)
        printf("Possible
");
    else
        printf("Impossible
");

    return 0;
}