HDU1247字典树

本题思路和一般字典树有所不同，将全部的字符串读入离线处理，insert操作和大部分字典树一致，接下来就是一个dfs函数，枚举全部字符串是否是符合条件的，dfs的tim参数表示了这个是第几次查询，如果遇到danger标记，就可以有两种路径，继续本次查询和进入下一个查询，这样讲每一个字符串都判断完毕即可。

 

我的代码：

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <algorithm>#include <stack>#include <bitset>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdlib>#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <ctime>#include <cmath>using namespace std;const int MAX=1000000;struct Node{Node* ne[26];bool danger;}node[MAX],*root;char in[50010][50];int K;Node* New(){Node* ret=&node[K++];for(int i=0;i<26;i++){ret->ne[i]=NULL;}ret->danger=false;return ret;}void init(){K=0;root=New();}void insert(char* s){char* p=s;Node* ptr=root;int id;while(*p){id=*(p++)-'a';if(ptr->ne[id]==NULL){ptr->ne[id]=New();}ptr=ptr->ne[id];}ptr->danger=true;}bool dfs(char* s,int tim){//tim=0 means that it is the first time search,//otherwise it is the second time.char* p=s;Node* ptr=root;int id;while(*p){id=*(p++)-'a';if(ptr->ne[id]==NULL){return false;//if the next pointer is null, return false.}ptr=ptr->ne[id];//important.//if this node is a danger case, we search a second time.if(ptr->danger){if(tim==0&&*p&&dfs(p,1)){return true;}}}//if the node is dangerous and the string has been completely search//return true.if(tim==1&&ptr->danger){return true;}else{return false;}}int main(){int n=0;init();while(gets(in[n])){insert(in[n]);n++;}for(int i=0;i<n;i++){if(dfs(in[i],0)){puts(in[i]);}}return 0;}