POJ 2513 Colored Sticks

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Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 36032		Accepted: 9427

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

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#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
int color, n;
int pre[500000+5];
int sum[500000+5];
struct node {
    node *next[26+5];
    bool flag;
    int id;
    node() {
        flag=0;
        id=0;
        for(int i=0;i<26;i++)
            next[i]=NULL;
    }
}*root;
int hash(char *s) {
    node *p=root;
    for(int i=0;s[i]!='\0';i++) {
        if(p->next[s[i]-'a']==NULL) {
            p->next[s[i]-'a']=new node;
        }
        p=p->next[s[i]-'a'];
    }
    if(p->flag) return p->id;
    else {
        p->flag=1;
        p->id=++color;
        return p->id;
    }
}
int fin(int x) {
    int r=x;
    while(x!=pre[x]) {
        x=pre[x];
    }
    while(r!=x) {
        int y=pre[r];
        pre[r]=x;
        r=y;
    }
    return x;
}
int main() {
    root=new node;
    n=color=0;
    memset(sum,0,sizeof(sum));
    for(int i=1;i<=250000;i++)
        pre[i]=i;
    char a[10+5], b[10+5], s[20+5];
    while(gets(s)&&strlen(s)){
        int i, j;
        for(i=0;s[i]!=' ';i++)
            a[i]=s[i];
        a[i]='\0';
        for(j=i+1;j<strlen(s);j++) {
            b[j-i-1]=s[j];
        }
        b[j-i-1]='\0';
        int num1=hash(a), num2=hash(b);
        n=max(num2,max(n,num1));
        pre[fin(num1)]=fin(num2);
        sum[num1]++;
        sum[num2]++;
    }
    int ans=0;
    for(int i=1;i<=n;i++)
        if(sum[i]%2) ans++;
    if(ans==0||ans==2) {
        int i;
        int x=fin(1);
        for(i=2;i<=n;i++) {
            if(fin(i)!=x)
               break;
        }
        if(i>n) cout<<"Possible"<<endl;
        else cout<<"Impossible"<<endl;
    }
    else cout<<"Impossible"<<endl;
}