POJ/PKU 2513 并查集+字典树+欧拉回路

 

Colored Sticks

Time Limit: 5000MS Memory Limit: 128000KTotal Submissions: 20750 Accepted: 5482

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

 

 

要仔细看题，木棍不仅仅只是可以一个方向放置，还可以倒过来，所以简单的说，这个题是一个无向图的欧拉回路判断

但是直接用map要超时。。我先开始就这样TLE了。后来看了下后面的discuss，发现要用字典树

于是又自己手动写了个字典树。

 

所以综上，我们需要：

1、字典树对每一个字符串进行标号

2、并查集判断整个图是否连通

3、判断无向图的欧拉回路是否存在

 

我的代码：

#include<stdio.h>#include<string.h>#include<algorithm>#define maxn 500005using namespace std;struct node{int num;node *next[26];void init();};node *root;int father[maxn];int temp[maxn];int sum[maxn];int cnt=0;void node::init(){this->num=0;}int find(int p){if(father[p]==p)return p;return father[p]=find(father[p]);}void Union(int a,int b){int x,y;x=find(a);y=find(b);if(x!=y)father[x]=y;}void init(){root= new node;int i;for(i=0;i<26;i++)root->next[i]=NULL;root->init();}int search(char *s){int i,len=strlen(s),j;node *p,*t;p=root;for(i=0;i<len;i++){if(p->next[s[i]-'a']==NULL){t=new node;for(j=0;j<26;j++)t->next[j]=NULL;t->init();p->next[s[i]-'a']=t;}p=p->next[s[i]-'a'];}if(p->num==0){cnt++;return p->num=cnt;}elsereturn p->num;}int main(){int i,a,b;char s1[15],s2[15];memset(sum,0,sizeof(sum));for(i=1;i<=maxn;i++)father[i]=i;init();while(scanf("%s%s",s1,s2)!=EOF){a=search(s1);b=search(s2);Union(a,b);sum[a]++;sum[b]++;}int NUM=0;for(i=1;i<=cnt;i++)if(father[i]==i)NUM++;if(NUM>1){printf("Impossible\n");return 0;}NUM=0;for(i=1;i<=cnt;i++)if(sum[i]&1)NUM++;if(NUM==0||NUM==2)printf("Possible\n");elseprintf("Impossible\n");return 0;}