POJ 2513 Colored Sticks(字典树+欧拉路径)

题目:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=11158

Colored Sticks
Time Limit: 5000MS   Memory Limit: 128000KB   64bit IO Format: %I64d & %I64u

Submit Status

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.
开始想用map<string,int>写的,但是string是cin输入啊,而hint的提示要用scanf。这可怎么办?后来发现了神奇的字典树,轻便高效值得拥有!映射的问题解决了,然后就是欧拉路径/回路的判断了。
相关知识点:
欧拉回路和欧拉路径的判断
欧拉回路:
无向图:每个顶点的度数都是偶数,则存在欧拉回路。
有向图:每个顶点的入度都等于出度,则存在欧拉回路。
欧拉路径:
无向图:当且仅当该图所有顶点的度数为偶数 或者 除了两个度数为奇数外其余的全是偶数。
有向图:当且仅当该图所有顶点 出度=入度 或者 一个顶点 出度=入度+1,另一个顶点 入度=出度+1,其他顶点 出度=入度。
本题和hdu 1116(上一篇博客)稍有不同,这是无向图,那个是有向图(单词不可能倒着写)。判定条件不要混淆。相关博客: http://blog.csdn.net/zzran/article/details/9106329
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=5e5+5;
struct node{
     int num;
     bool is_word;
     node *next[26];
     node(){
          num=is_word=0;
          memset(next,NULL,sizeof(next));
     }
};
node *root;
int n,c[maxn],f[maxn];
void insert(char s[]){
     node *p=root;
     for(int i=0;s[i];i++){
          int dex=s[i]-'a';
          if(p->next[dex]==0)  p->next[dex]=new node();
          p=p->next[dex];
     }
     if(p->is_word==0){ // important!  because of num++
         p->num=++n;  //num is dex of string
         p->is_word=1;
     }
}
int find(int x){
     if(x==f[x]) return x;
     f[x]=find(f[x]);
     return f[x];
}
int getdex(char s[]){
     node *p=root;
     for(int i=0;s[i];p=p->next[s[i]-'a'],i++);  // not need length
     return  p->num;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    n=0;
    memset(c,0,sizeof(c));
    memset(f,0,sizeof(f));
    root=new node();
    char a[15],b[15];
    int adex,bdex;
    while(~scanf("%s%s",a,b)){
         insert(a);
         insert(b);
         adex=getdex(a);
         bdex=getdex(b);
         c[adex]++;  //all count
         c[bdex]++;
         if(!f[adex]) f[adex]=adex;  // U gather。
         if(!f[bdex]) f[bdex]=bdex;
         f[find(adex)]=f[find(bdex)];
    }
    int oddsum=0;
    bool judge=1;
    for(int i=1;i<=n;i++){
        if(c[i]&1) oddsum++;
        if(find(i)!=f[1]){
            judge=0;
            break;
        }
    }
    if((oddsum==0 || oddsum==2)&&judge) puts("Possible");
    else puts("Impossible");
    return 0;
}


你可能感兴趣的:(poj,欧拉路径)