POJ 2513 Colored Sticks(欧拉回路判断+字典树Trie+并查集)

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 red red violet cyan blue blue magenta magenta cyan Sample Output Possible Hint Huge input,scanf is recommended. Source ​​The UofA Local 2000.10.14​​

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 40272		Accepted: 10463

题意：

给一些木棒，木棒两端都涂上颜色，求是否能将木棒首尾相接，连成一条直线，要求不同木棒相接的一边必须是相同颜色的

分析：

典型的判断欧拉路和欧拉回路问题，关键映射map竟然超时，用字典树就过来

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
using namespace std;
const int N=260000;
 
int n, m;
int f[N*2],degree[N*2];//记录第i点的度数
const int maxnode=5000000+1000;
const int sigma_size=26;
int cnt;//记录当前已经读入了cnt个不同的颜色了
struct Trie
{
    int ch[maxnode][sigma_size];
    int val[maxnode];//单词节点的val才非0,且val[i]表示的是该单词节点的编号
    int sz;
    void init()
    {
        sz=1;
        memset(ch[0],0,sizeof(ch[0]));
        val[0]=0;
    }
    int insert(char *s)
    {
        int n=strlen(s),u=0;
        for(int i=0;i<n;i++)
        {
            int id=s[i]-'a';
            if(ch[u][id]==0)
            {
                ch[u][id]=sz;
                val[sz]=0;
                memset(ch[sz],0,sizeof(ch[sz]));
                sz++;
            }
            u=ch[u][id];
        }
        if(val[u]==0)//新颜色
            val[u]=++cnt;
        return val[u];
    }
}trie;
int find_(int x)
{
    if(f[x]==-1)    return x;
    return f[x] = find_(f[x]);
}
void merge_(int x, int y)
{
    int t1, t2;
    //cout<<f[x]<<endl;
    t1 = find_(x); 
    t2 = find_(y);
    if (t1 != t2)   f[t2] = t1;
    else return;
}
int isEuler()
{
    int cnt=0;
    for (int i = 1; i <= n; i++)
        {
            if (degree[i] & 1)  
            cnt++;
            if(cnt>2) return 0;
        }
        
    if(cnt==2||cnt==0) return 1;
    return 0;
}
int isconnect()
{
    int cnt = 0;
    int ff=find_(1);
    for (int i=2; i <= n; i++)
    {
        if (find_(i)!=ff)
        return 0;
    }
    return 1;
}
int main()
{
        memset(f,-1,sizeof(f));
        memset(degree, 0, sizeof(degree));
        cnt=0;
        trie.init();
        char t1[100],t2[100];
        n=1;
        //int i=0;
        while(scanf("%s%s",t1,t2)!=EOF)
        {
            int i=trie.insert(t1);//尝试插入新颜色
            int j=trie.insert(t2);
        
            
            //cout<<mapp[t1]<<" "<<mapp[t2]<<endl;
            degree[i]++;
            degree[j]++;
            
            merge_(i,j);
              //i++;
            //if(i==5) break;
        }
        n=cnt;
        //cout<<"sss"<<endl;
        if(isconnect()&&isEuler())  
        printf("Possible");
        else
        printf("Impossible");
    return 0;
}