poj 2513 （并查集与字典树）

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 29235		Accepted: 7710

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

题目意思就是说能不能把每个stick都连接起来成一行（即能不能构成一个欧拉路baike.baidu.com/view/566040.htm?fr=aladdin），能就Possible，反之Impossible。

我原来打算是用map做的，结果超时了，后来打算用哈希函数确定id，结果找不到合适的函数。无奈只能用trie了。

首先用trie确定每种颜色的id，然后确定每种颜色出现次数，欧拉判断一下，再并查集一下就行了。

AC代码：

#include<iostream>
#include<map>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int f[1500];
int a[1500];
int trie[510000][26],num[510000][26];
int n,r;
int insert(char *ch){
    int j=0;
    while(*ch){
        if(trie[j][*ch-'a']) j=trie[j][*ch-'a'];
        else{
            ++r;
            trie[j][*ch-'a']=r;
            j=r;
        }
        ch++;
    }
    ch--;
    if(!num[j][*ch]) num[j][*ch]=++n;    //确定该字符串是第几个字符 
    return num[j][*ch];     //返回该字符串的id 
}
int find(int x){          //并查集 
    if(x==f[x]) return x;
    return f[x]=find(f[x]);
}
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    memset(a,0,sizeof(a));
    memset(trie,0,sizeof(trie));
    memset(num,0,sizeof(num));
    n=r=0;
    for(int i=1;i<1500;i++)      //初始化并查集 
        f[i]=i;
    char s1[15],s2[15];
    int x,y;
    while(scanf("%s%s",s1,s2)!=EOF){   //数据很多，建议用scanf，cin可能超时 
        x=insert(s1);
        y=insert(s2);
        a[x]++; a[y]++;
        x=find(x);
        y=find(y);
        if(x!=y) 
            f[x]=y;
    }
    int cnt=0;
    for(int i=1;i<=n;i++){
        if(f[i]==i) cnt++;
        if(cnt>=2)
            break;
    }
    if(cnt>=2) cout<<"Impossible"<<endl;        //并查集判断是否都属于一个集合 
    else{
        int sum=0;
        for(int i=1;i<=n;i++){
            if(a[i]&1) 
                sum++;
            if(sum>2) break;
        }
        if(sum!=0 && sum!=2)                  //欧拉判断是否构成欧拉路 
                cout<<"Impossible"<<endl;
        else
            cout<<"Possible"<<endl;
    }
    return 0;
}