POJ2513 Colored Sticks (并查集+trie)

传送门

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K

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.

Source

The UofA Local 2000.10.14

 

回顾了一下欧拉图的知识。。整个图的奇度点个数如果>=3或只有1个 就不能一笔画画完

中间re了几次。。因为trie里我开了n*30，但是外面数组只开了n...

#include<set>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 250010;
#define For(i,n) for(int i=1;i<=n;i++)
#define For0(i,n) for(int i=0;i<n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)

char s1[11],s2[11];
int n,in[N*30],fa[N*30],ans;
struct Trie{
    int sz,ch[N*30][26],v[N*30],Loc[N*30];
    Trie(){sz=1;}
    int insert(char st[]){
        int now=1;v[1]++;int len=strlen(st);
        For0(i,len){
            int id=st[i]-'a';
            if(!ch[now][id]) ch[now][id]=++sz;
            v[now=ch[now][id]]++;
        }
        if(Loc[now]) return Loc[now];
        else         return Loc[now]=++n;
    }
}trie;

int find(int i){
    if(fa[i]==i) return i;
    else         return fa[i]=find(fa[i]);
}

int main(){
    #ifndef ONLINE_JUDGE
        freopen("trie.in","r",stdin);
    #endif // ONLINE_JUDGE
    For0(i,N*30) fa[i]=i;
    while(scanf("%s %s",&s1,&s2)!=EOF){
        int x=trie.insert(s1),y=trie.insert(s2);
        in[x]++;in[y]++;
        int fx=find(x),fy=find(y);
        if(fx!=fy) fa[fx]=fy;
    }
    int checks=find(1);
    Rep(i,2,n) if(find(i)!=checks){
        puts("Impossible\n");
        return 0;
    }
    For(i,n) {
        in[i]=in[i]%2;
        if(in[i]) ans++;
    }
    if(ans==1||ans>=3) puts("Impossible");
    else               puts("Possible");
    return 0;
}