Colored Sticks---POJ2513

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 39360		Accepted: 10260

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.

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <string.h>
#include <queue>
#include <map>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <string.h>
#include <queue>
#include <map>
int trie[500000][26],flag[500000],r[500000],pre[500000],num[500000],s,n;
char ss[15];
void insert(int w,int k){
	++s;
    flag[s] = 0;
    memset(trie[s], 0, sizeof(trie[s]));
    trie[w][k]=s;
}
int add(char we[]){
	int c=0;
	int len=strlen(we);
	for(int i=0;i<len;i++){
		int k=we[i]-'a';
		if(trie[c][k]==0){
			insert(c,k);
		}
		c=trie[c][k];
	}
    if(flag[c])
        ++ num[r[c]];
    else
    {
        flag[c] = 1;
        r[c]=++n;
        num[n] = 1;
    }
    return r[c];
}
int find(int x){
	if(pre[x]==x)
		return x;
	return pre[x]=find(pre[x]);
}
int main(){
	char a[15],b[15];
	for(int i=0;i<500000;i++)
		pre[i]=i;
	n=0;
	s=0;
	memset(num, 0, sizeof(num));
	while(scanf("%s %s",a,b)!=EOF){
		int x=add(a);
		int y=add(b);
        int tx = find(x), ty = find(y);
        if(tx != ty)
            pre[tx] = ty;
	}
	int gg=find(1);
	int f=0,k=0;
	for(int i=2;i<=n;i++){
		if(find(i)!=gg){
			f=1;
			break;
		}
        if(num[i] % 2)
            ++k;
        if(k>2){
        	f=1;
        	break;
		}
	}
	if(f==1){
		printf("Impossible\n");
	}
	else {
		printf("Possible\n");
	} 
}