Poj 2513 Colored Sticks//Trie+並查集+歐拉通路

  這道題可以直接看出是用歐拉通路來做，但是資料特別大，只能想辦法壓縮資料。所以用Trie樹來儲存幾萬種顏色，並給這些顏色標記為相應的結點。

  下面是代碼：

#include<stdio.h>#include<string.h>#define maxn 510000int ch[maxn][26];int val[maxn];int degree[maxn];int f[maxn];int cnt;int ID;char s1[11],s2[11];void initi(){    for(int i = 0; i < maxn; i++) f[i] = i;    memset(ch[0],0,sizeof(ch[0]));    ID = 0;    cnt = 1;}int find_father(int x){    if(x != f[x])    {        f[x] = find_father(f[x]);    }    return f[x];}void union_node(int x,int y){    x = find_father(x);    y = find_father(y);    if(x != y) f[x] = y;}void insert_Trie(char *s){    int u = 0;    while(*s)    {        int id = *s - 'a';        if(ch[u][id] == 0)        {            memset(ch[cnt],0,sizeof(ch[cnt]));            val[cnt] = -1;            ch[u][id] = cnt++;        }        u = ch[u][id];        s++;    }    val[u] = ID;}int get_id(char *s){    int u = 0;    while(*s)    {        int id = *s - 'a';        if(ch[u][id] == 0) return -1;        u = ch[u][id];        s++;    }    if(val[u] >= 0) return val[u];    return -1;}void solve(){    bool ok = true;    int i;    for(i = 0; !degree[i]; i++);    for(int j = i + 1; j < ID; j++)    {        if(degree[j] && find_father(j) != find_father(i))        {            ok = false;            break;        }    }    int flag = 0;    if(ok)    {        for(int j = 0; j < ID; j++)        {            if(degree[j] %2 != 0) flag++;            if(flag > 2)            {                ok = false;                break;            }        }    }    if(ok && (flag == 2 || flag == 0) ) printf("Possible\n");    else printf("Impossible\n");}int main(){        initi();        while(scanf("%s %s",s1,s2) == 2)        {            int u = get_id(s1);            int v = get_id(s2);            if(u == - 1)  insert_Trie(s1),u = ID++;            if(v == -1) insert_Trie(s2),v = ID++;            union_node(u,v);            degree[u]++;            degree[v]++;        }        solve();       return 0;}