Trie树+并查集+欧拉回路poj2513

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Colored Sticks

Time Limit: 5000MS Memory Limit: 128000KTotal Submissions: 30307 Accepted: 8007

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

题意：有很多木棒，每根木棒两端有颜色，颜色相同的可以接起来，问可不可以一笔画

思路：就跟单词接龙一样，每种颜色看成一个节点，每根木棒那个看成一条边，问题就转化成了是否存在欧拉回路。对单词进行编号时用Trie进行优化，并查集判断是不是联通

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=250010;char s1[100],s2[100];int in[maxn*2];int pre[maxn*2],id,num;struct node{    node *next[30];    int val;    node(){memset(next,0,sizeof(next));val=0;}};struct TREE{    node *root;    void clear(){root=new node();}    int idx(char x){return x-'a';}    int insert(char *s)    {        int n=strlen(s);        node *p=root;        for(int i=0;i<n;i++)        {            int c=idx(s[i]);            if(!p->next[c])                p->next[c]=new node();            p=p->next[c];        }        if(!p->val)p->val=num++;        return p->val;    }}tree;int find(int x){    if(x==pre[x])return x;    return pre[x]=find(pre[x]);}void unite(int a,int b){    int x=find(a),y=find(b);    if(x==y)return;    pre[y]=x;}int main(){    memset(in,0,sizeof(in));    num=1;    tree.clear();    for(int i=0;i<maxn*2;i++)pre[i]=i;    while(scanf("%s%s",s1,s2)!=EOF)    {        int x=tree.insert(s1);        int y=tree.insert(s2);        in[x]++,in[y]++;        unite(x,y);    }    int cnt=0;    for(int i=1;i<num;i++)        if(pre[i]==i)cnt++;    if(cnt>1){printf("Impossible\n");return 0;}    cnt=0;    for(int i=1;i<num;i++)        if(in[i]%2)cnt++;    if(cnt==2||cnt==0)printf("Possible\n");    else printf("Impossible\n");    return 0;}