【字典树】hdu 1247 Hat’s Words

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http://acm.hdu.edu.cn/showproblem.php?pid=1247

第二道字典树。

先建树，然后对每个词进行查询，若串存在前缀和后缀都属于整棵字典树，则满足题意

/*   hdu 1247 字典树   此时，cnt表示以该节点到根节点组成的串的个数，而非前缀的个数（区别上一题）   注意exactly 2的情况*/#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<algorithm>#include<sstream>#define eps 1e-9#define pi acos(-1)#define INF 0x7fffffff#define inf -INF#define MM 12900#define N 50using namespace std;typedef long long ll;const int _max = 5e4 + 10;struct Trie{  Trie *next[26];  int cnt;//从根节点到该节点组成的串为前缀的个数}*root;void insert(char *s){  Trie *p = root,*pnew;  for(int i = 0; i < strlen(s);++ i){    int x = s[i]-'a';    if(p->next[x]==NULL){       pnew = new Trie;       pnew->cnt = 0;       for(int j = 0; j < 26; ++ j)            pnew->next[j] = NULL;       p->next[x] = pnew;    }    //else p->next[x]->cnt++;    p=p->next[x];  }  p->cnt = 1;}int search(char *s){  if(*s=='\0') return 0;  Trie *p = root;  for(;*s!='\0';++s){    int x = *s - 'a';    if(p->next[x]==NULL) return 0;    p=p->next[x];  }  if(p->cnt==1)return 1;//exactly 2  return 0;}int judge(char *s){  Trie *p = root;  for(;*s!='\0';++s){    int x = *s - 'a';    if(p->next[x]){        p = p->next[x];        if(p->cnt==1&&search(s+1)) return 1;//关键！    }  }  return 0;}void init(){  root = new Trie;  root->cnt = 0;  for (int i = 0;i < 26;i ++)    root->next[i] = NULL;}int main(){#ifndef ONLINE_JUDGE    freopen("input.txt","r",stdin);#endif // ONLINE_JUDGE    char str[_max][20];//假定字符串长度不超20    init();    int i = 0;    while(gets(str[i])){      insert(str[i]);++i;    }    for(int j = 0; j<i;++ j){        if(judge(str[j])) printf("%s\n",str[j]);    }    return 0;}

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