nyoj 230 涂彩棒 并查集 字典树 欧拉

彩色棒

时间限制： 1000 ms  |  内存限制： 128000 KB

难度： 5

描述

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?

输入

the frist line have a number k(0<k<=10),that is the number of case. each case contais a number m,then have m line, 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 13 characters. There is no more than 250000 sticks.

输出

If the sticks can be aligned in the desired way, output a single line saying Possible, otherwise output Impossible.

样例输入

1

5

blue red

red violet

cyan blue

blue magenta

magenta cyan

样例输出

Possible

#include<stdio.h>

#include<string.h>

#include<malloc.h>

struct node

{

    int order;

    struct node *next[26];

};



int  t;//记录颜色种类

int father[250001];//记录并查集父亲节点

int degree[250001];//记录每种颜色出现的次数

int insert(char *str,node *T)//插入构建字典树

{

    int len,id,flat=0,i,j;

    node *p,*q;

    len=strlen(str);

    p=T;

    for(i=0;i<len;++i)

    {

        id=str[i]-'a';

        if(p->next[id]==NULL)

        {

            flat=1;

            q=(node *)malloc(sizeof(node));

            for(j=0;j<26;++j)

                q->next[j]=NULL;

            p->next[id]=q;

        }

        p=p->next[id];

    }

    if(flat)//字典树向前延伸，出现新的颜色

    {

        p->order=t;//在此记录此种颜色的 序号

        degree[t++]++;//增加出现次数

        return p->order;

    }

    else

    {

        if(p->order==0)//可能有这种情况 abdse abd虽然没有延伸，但也是新颜色

        {

            p->order=t;

            degree[t++]++;

            return p->order;

        }

        degree[p->order]++;

        return p->order;

    }

}



int findfather(int i)

{

    if(father[i]==-1)

        return i;

    else

        return findfather(father[i]);

}



void merge(int a,int b)//合并 

{

    a=findfather(a);

    b=findfather(b);

    if(a!=b)

    {

        if(father[a]<father[b])

            father[b]=a;

        else

            father[a]=b;

    }

}



int main()

{

    char str1[14],str2[14];

    int flag,i,num1,num2,n,x;

    node *T;

    scanf("%d",&x);

    while(x--)

    {

        T=(node *)malloc(sizeof(node));

        T->order=0;

        memset(degree,0,sizeof(degree));

        memset(father,-1,sizeof(father));

        for(i=0;i<26;++i)

            T->next[i]=NULL;

        t=1;

        scanf("%d",&n);

        if(n==0)

        {

            printf("Possible\n");

            continue;

        }

        while(n--)

        {

            scanf("%s%s",str1,str2);

            num1=insert(str1,T);

            num2=insert(str2,T);

            merge(num1,num2);

        }

        flag=0;

        for(i=1;i<t;++i)

            if(father[i]==-1)//统计根节点

                ++flag;

        if(flag>1)

            printf("Impossible\n");

        else

        {

            flag=0;

            for(i=1;i<t;++i)

                if(degree[i]&1)

                    flag++;

            if(flag==2||flag==0)

                 printf("Possible\n");

            else

                 printf("Impossible\n");

        }

    }

    return 0;

}







            

泪奔啊，刚开始想着和单词连接很像，就把那个代码作了一点改动，一直超时，后来 胖子说他做字典树用的并查集。并查集对于大数据量有很快的效率，这个题是25万的数据，看来就是考这点啦。用了并查集，果然过啦

思路：

先把数据存储到字典树里面，字典树再这路的作用有两个：一个是统计颜色种类，一个是每种颜色数量。然后用并查集，讲这些颜色合并。每次输入都合并。到最后统计下 根节点的数量，就是并查集的数量，如果大于1，说明不单单在一个集合，绝对不能连成线，如果是，则要根据 欧拉通路的原理去判定