POJ 2513 Colored Sticks

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 redred violetcyan blueblue magentamagenta cyan

Sample Output

Possible

Hint

Huge input,scanf is recommended.

Source

The UofA Local 2000.10.14

/*

     题目大意:

            给定不超过250000根棍子，棍子上平分为分为两种颜色，问可不可以将所有的棍子(连接点颜色相同)连在一起，如果可以的话输出Possible否则Impossible.

     解题思路:

            将题目理解为给定一些无向边，将颜色理解为一个个的点，题目就转化为求这些无向边中是否存在一条欧拉路径，将所有的边全都经过而不重复。关键是将颜色转换为独立点，需要用到字典树(trie)，刚刚看到这道题想用map，不过看到250000的数据就没敢...用map映射的话应该会超的，这时从POJ 2418上学习 到的经验！剩下的就好理解了，用字典树返回颜色编号，用并查集判断这个图是否为联通图，不是联通图不会有欧拉回路存在。最后通过每个点的度数判断是否为欧拉路或欧拉回路就ok了。记得数组至少开到500001，有边界数据(亲测)，刚开始还以为250000根棍子最多有125000种颜色，开到126000足够了，结果RE。

*/

#include <iostream>#include <algorithm>#include <string>#include <cstring>#include <queue>#include <stack>#include <cmath>#include <vector>#include <cstdio>#include <map>#include <set>const int INF = 0x3f3f3f3f;using namespace std;/*代码较挫，字典树虽说拿手但是我写的一般都点乱*/struct node{    int num;    node *next[26];}*head,*p,*q;int i,k = 1,flag,ant,ans[500001],vis[500001];int g(int x){    while(x != vis[x])        x = vis[x];    return x;}void f(int x,int y){    int u = g(x);    int v = g(y);    if(u != v)    {        vis[u] = v;    }}node* getnew(node *l){    l = new node;    for(int i=0; i<26; i++)        l->next[i] = NULL;    l->num = -1;    return l;}void gettree(char stu[]){    for(i = 0,p = head; stu[i]!='\0' ; i++)    {        if(stu[i+1] == '\0')        {            if(p->next[stu[i]-'a'] == NULL)            {                q = getnew(q);                q->num = k++;                p->next[stu[i]-'a'] = q;                if(flag == 1)                    ant = q->num;                if(flag == 2)                {                    f(ant,q->num);ans[ant]++;ans[q->num]++;                }            }            else            {                if(p->next[stu[i]-'a']->num == -1)                    p->next[stu[i]-'a']->num = k++;                if(flag == 1)                    ant = p->next[stu[i]-'a']->num;                if(flag == 2)                {                    f(ant,p->next[stu[i]-'a']->num);ans[ant]++;ans[p->next[stu[i]-'a']->num]++;                }            }        }        else        {            if(p->next[stu[i]-'a'] == NULL)            {                q = getnew(q);                p->next[stu[i]-'a'] = q;                p = p->next[stu[i]-'a'];            }            else                p = p->next[stu[i]-'a'];        }    }}int main(){    char stu[15],stv[15];    for(i=0;i<=500000;i++)    {        vis[i] = i;ans[i]= 0;    }    head = getnew(head);    while(~scanf("%s %s",stu,stv))    {        flag = 1;gettree(stu);flag = 2;gettree(stv);    }    bool Flag = false,F = false;    for(i = 1;i < k;i++)    {        if(vis[i] == i && !Flag)        {            Flag = true;        }        else if(vis[i] == i && Flag)        {            F = true;break;        }    }    if(F)        printf("Impossible\n");    else    {        int s = 0;        for(i = 1;i < k;i++)        {            if(ans[i]%2)                s++;        }        if((s == 2) || (s == 0))            printf("Possible\n");        else            printf("Impossible\n");    }    return 0;}