poj2513——判断是否为存在欧拉路径的连通图（并查集，trie树）

POJ 2513    Colored Sticks

欧拉回路判定，并查集，trie树

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 31621		Accepted: 8370

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：给若干根两端涂上不同颜色的筷子，相同颜色的筷子端点可接在一起，问是否能将筷子接成一条直线
思路：将每种颜色看成结点，筷子看成边，建无向图，判断无向图是否是存在欧拉路径的连通图
判断欧拉图：用并查集判断图是否连通，判断是否存在欧拉回路，不存在奇点
判断存在欧拉路径：存在0或两个奇点

难点：题目卡了map和hash。。对字符串的转化只能用trie树了。。。

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<map>

#include<vector>



using namespace std;



const int maxn=510010;



int cnt=0;

int indeg[maxn];

int fa[maxn];



struct Node

{

    int id;

    Node *next[26];

};Node root;



int insert(char*s)   //返回结点的id

{

    Node *p=&root;

    for(int i=0;i<strlen(s);i++){

        if(p->next[s[i]-'a']==NULL){

            Node *newnode=(Node*)malloc(sizeof(Node));

            memset(newnode,0,sizeof(Node));

            p->next[s[i]-'a']=newnode;

        }

        p=p->next[s[i]-'a'];

    }

    if(p->id) return p->id;  //如果结点已存在，直接返回

    return p->id=++cnt;   //不存在则返回新结点

}



int find(int x)

{

    return fa[x]==x?x:fa[x]=find(fa[x]); 

}



int main()

{

    char s1[20],s2[20];

    memset(indeg,0,sizeof(indeg));

    for(int i=0;i<maxn;i++) fa[i]=i;

    char s[40];

    while(gets(s)&&strlen(s)){

        sscanf(s,"%s%s",s1,s2);

        int u=insert(s1),v=insert(s2);

        indeg[u]++;indeg[v]++;    //记录入度

        int x=find(u),y=find(v);    

        if(x!=y) fa[x]=y;     //记录合并连通分量

    }

    int n=0;

    bool flag=1;

    for(int i=1;i<=cnt;i++){

        if(indeg[i]&1) n++;

        if(n>2){      //判断奇点数目

            flag=0;break;

        }

        if(find(i)!=find(1)){  //判断连通性

            flag=0;break;

        }

    }

    if(flag) cout<<"Possible"<<endl;

    else cout<<"Impossible"<<endl;

    return 0;

}

poj2513