POJ 3283 典型的字典树trie tree 问题

来源：互联网发布：淘宝上有卖A片吗编辑：程序博客网时间：2024/06/05 16:07

本题是要实现一个倒置的字典树，此处用map来存储键值key以及所对应的权值value，方便搜索；

本题有一个需要注意的地方；由于要从后开始向前来建立字典树，尽量使用栈 stack 来存储 key 值；

而不要获取长串在解析串，解析的时候很难解析对，因为里边或是夹杂多个“空格”，获取其他的字符，

考虑不全不可能解出本题，所以要避免；

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <map>#include <stack>#include <memory.h>using namespace std;#define N 100010std::map<string,int> HASH;std::stack<string> sta;string card;int cs,num,i,k,len,pos;int idn;    //标志当前节点的个数struct node{    struct node * next[53];    void init()    {        for( i=0 ;i<53;i++)            next[i] = NULL;    }} *root;void inithash(){    HASH["AC"] = 0; HASH["2C"] = 1; HASH["3C"] = 2; HASH["4C"] = 3;    HASH["5C"] = 4; HASH["6C"] = 5; HASH["7C"] = 6; HASH["8C"] = 7;    HASH["9C"] = 8; HASH["10C"] = 9; HASH["JC"] = 10; HASH["QC"] = 11;    HASH["KC"] = 12; HASH["AD"] = 13; HASH["2D"] = 14; HASH["3D"] = 15;    HASH["4D"] = 16; HASH["5D"] = 17; HASH["6D"] = 18; HASH["7D"] = 19;    HASH["8D"] = 20; HASH["9D"] = 21; HASH["10D"] = 22; HASH["JD"] = 23;    HASH["QD"] = 24; HASH["KD"] = 25; HASH["AH"] = 26; HASH["2H"] = 27;    HASH["3H"] = 28; HASH["4H"] = 29; HASH["5H"] = 30; HASH["6H"] = 31;    HASH["7H"] = 32; HASH["8H"] = 33; HASH["9H"] = 34; HASH["10H"] = 35;    HASH["JH"] = 36; HASH["QH"] = 37; HASH["KH"] = 38; HASH["AS"] = 39;    HASH["2S"] = 40; HASH["3S"] = 41; HASH["4S"] = 42; HASH["5S"] = 43;    HASH["6S"] = 44; HASH["7S"] = 45; HASH["8S"] = 46; HASH["9S"] = 47;    HASH["10S"] = 48; HASH["JS"] = 49; HASH["QS"] = 50; HASH["KS"] = 51;}void build( ){    struct node * temp = root;    int curid;    while( !sta.empty() )    {        card = sta.top();        sta.pop();        curid = HASH[card];        if( temp->next[curid] != NULL)            temp = temp->next[curid];        else        {            temp->next[curid] = new struct node;            temp = temp->next[curid];            temp->init();            idn++;        }    }}void del( struct node * temp){    if(temp == NULL)        return;    for( i=0;i<53;i++)    {        if(temp->next[i])            del(temp->next[i]);    }    delete temp;    temp = NULL;    return;}int main(){    inithash();    while(cin>>cs && cs)    {        root = new struct node;        root->init();        idn = 0;        while( !sta.empty() )            sta.pop();        for( k=1 ;k <= cs ;k++)        {            cin>>num;            int j;            for(j=1;j<=num;j++)            {                cin>>card;                sta.push(card);            }            build();        }        cout<<idn<<endl;        del(root);    }    return 0;}