poj 2513 Colored Sticks (字典树+欧拉回路判定)

Colored Sticks

Time Limit: 5000MS Memory Limit: 128000KTotal Submissions: 27792 Accepted: 7346

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

感想：

以为这题很水，开始写的两个代码果断TLE，后来才发现要用字典树的知识，然后又是WA，原来还要考虑输入为空的时候的情况要输出“Possible”，汗。。。！这数据，真蛋疼。还有要注意的是这题应该看做无向图。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <map>#define maxn 500005#define MAX 26using namespace std;int n,m,ans,t,num;int sset[maxn],srank[maxn];int in[maxn];char s1[105],s2[105];struct Trie                   //Trie结点声明{    int isStr;               //标记该结点处是否构成单词    Trie *next[MAX];          //儿子分支};void insert(Trie *root,const char*s)    //将单词s插入到字典树中{    if(root==NULL||*s=='\0') return;    int i;    Trie *p=root;    while(*s!='\0')    {        if(p->next[*s-'a']==NULL)       //如果不存在，则建立结点        {            Trie *temp=(Trie *)malloc(sizeof(Trie));            for(i=0; i<MAX; i++)            {                temp->next[i]=NULL;            }            temp->isStr=0;            p->next[*s-'a']=temp;            p=p->next[*s-'a'];        }        else        {            p=p->next[*s-'a'];        }        s++;    }    p->isStr=++t;                  //单词结束的地方标记此处可以构成一个单词}int search(Trie *root,const char*s) //查找某个单词是否已经存在{    Trie *p=root;    while(p!=NULL&&*s!='\0')    {        p=p->next[*s-'a'];        s++;    }    if(p!=NULL&&p->isStr!=0) return p->isStr;  //在单词结束处的标记为true时，单词才存在    return 0;}void del(Trie *root)                     //释放整个字典树占的堆区空间{    int i;    for(i=0; i<MAX; i++)    {        if(root->next[i]!=NULL)        {            del(root->next[i]);        }    }    free(root);}void init(){    int i;    for(i=0; i<maxn; i++)    {        sset[i]=i;        srank[i]=1;    }    num=0;}int sfind(int x)        // 查找时优化了路径{    int r=x,nn;    while(sset[r]!=r) r=sset[r];    while(x!=r)    {        nn=sset[x];        sset[x]=r;        x=nn;    }    return r;}void smerge(int a,int b)    // 合并时将深度小的集合合并到大的里面{    int x,y;    x=sfind(a);    y=sfind(b);    if(x!=y)    {        if(srank[x]<srank[y])  sset[x]=y;        else        {            sset[y]=x;            if(srank[x]==srank[y])  srank[x]++;        }        num++;    }}bool solve(){    int i,j,u,h;    u=0;    if(num!=t-1) return false ;    for(i=1;i<=t;i++)    {        if(in[i]&1) u++;    }    if(u==0||u==2) return true ;    return false ;}int main(){    int i,j,x,y;    t=0;    init();    Trie *root= (Trie *)malloc(sizeof(Trie));    for(i=0; i<MAX; i++)    {        root->next[i]=NULL;    }    root->isStr=0;    memset(in,0,sizeof(in));    while(~scanf("%s%s",s1,s2))    {        x=search(root,s1);        if(!x)  // 不存在就建立        {            insert(root,s1);            x=t;        }        y=search(root,s2);        if(!y)        {            insert(root,s2);            y=t;        }        in[x]++;        in[y]++;        smerge(x,y);    }    if(t==0)    {        printf("Possible\n");        return 0;    }    if(solve()) printf("Possible\n");    else printf("Impossible\n");//    del(root);    return 0;}