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poj2513Colored Sticks(欧拉通路+字典树+并查集)

2018年02月22日 ⁄ 综合 ⁄ 共 1672字 ⁄ 字号 评论关闭

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.
解题:欧拉通路定义:在一个连通图G中,每条边只经过一次,一次性走完所有的边。
    欧拉通路判断:每个顶点的度数全为偶数或恰好只有两个顶点的度数为奇数。
那么这题要是接成一条线,也就是每条边只用一次,正合欧拉通路定义。 用并查集来判断所有的边是不是组成一个连能图,字典树
来计算顶点的个数及每个顶点出现的次数(计算奇偶数)。
注意:这题数据有空输入,那么也是possible.
#include<stdio.h>
typedef struct nn
{
    int flag;
    int indx;
    struct nn *next[26];
}node;
node *builde()
{
    node *p=new node;
    p->flag=0;
    for(int i=0;i<26;i++)
    p->next[i]=NULL;
    return p;
}
node *root;
int jdn;//计算顶点为奇数的个数
int insert(char str[],int n)
{
    node *p=root;
    for(int i=0;str[i]!='\0';i++)
    {
        if(p->next[str[i]-'a']==NULL)
        p->next[str[i]-'a']=builde();
        p=p->next[str[i]-'a'];
    }
    if(p->flag==0)
    p->indx=n;
    p->flag++;
    if((p->flag)%2)jdn++;else jdn--;
    return p->indx;
}

int fath[5100000],dn;//dn为边数和点数之差
int findfather(int x)
{
    if(x!=fath[x])
    fath[x]=findfather(fath[x]);
    return fath[x];
}
void set(int a,int b)
{
    a=findfather(a);
    b=findfather(b);
    if(a!=b)
    {
        dn--; fath[a]=b;
    }
}
int main()
{
    char c1[1000],c2[1000];
    int n=0,a,b;
    dn=0; jdn=0;
    root=builde();
    while(scanf("%s%s",c1,c2)>0)
    {
        a=insert(c1,n+1);
        if(a==n+1)
        {
            dn++; n++; fath[n]=n;
        }
        b=insert(c2,n+1);
        if(b==n+1)
        {
            dn++; n++; fath[n]=n;
        }
        set(a,b);
    }
    if((jdn==2||jdn==0)&&dn==1||n==0)
    printf("Possible\n");
    else
    printf("Impossible\n");
}

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