【練習04】 字典樹 1006 統計難題

這一題數組要開到百萬級，如1000005，不然會RE。

解題思路：字典樹，為了最後統計出首碼被包含的次數，需要在構造trie的時候增加一個數組記錄經過每個節點的路徑數。

//模板開始#include <string>   #include <vector>   #include <algorithm>   #include <iostream>   #include <sstream>   #include <fstream>   #include <map>   #include <set>   #include <cstdio>   #include <cmath>   #include <cstdlib>   #include <ctime>#include<iomanip>#include<string.h>#define SZ(x) (int(x.size()))using namespace std;int toInt(string s){istringstream sin(s); int t; sin>>t; return t;}template<class T> string toString(T x){ostringstream sout; sout<<x; return sout.str();}typedef long long int64;int64 toInt64(string s){istringstream sin(s); int64 t; sin>>t;return t;}template<class T> T gcd(T a, T b){ if(a<0) return gcd(-a, b);if(b<0) return gcd(a, -b);return (b == 0)? a : gcd(b, a % b);}#define ifs cin//模板結束（通用部分）#define maxnode 1000005#define sigma_size 26#define str_len 15int ch[maxnode][sigma_size];int ch_count[maxnode];int val[maxnode];struct Trie{int sz;Trie(){sz = 1;memset(ch[0], 0, sizeof(ch[0]));memset(ch_count, 0, sizeof(ch_count));}int idx(char c){return c - 'a';}void insert_trie(char* s, int v){int u = 0, n = strlen(s);for(int i = 0; i < n; i++){int c = idx(s[i]);if(!ch[u][c]){memset(ch[sz], 0, sizeof(ch[sz]));val[sz] = 0;ch[u][c] = sz++;}u = ch[u][c];ch_count[u]++;}val[u] = v;}int find_trie(char* s){int u = 0;int n = strlen(s);int flag = 1;for(int i = 0; i < n; i++){int c = idx(s[i]);if(!ch[u][c]){flag = 0;break;}u = ch[u][c];}if(flag == 0){return 0;}else{return ch_count[u];}}};//hdoj 1251 統計難題int main(){//ifstream ifs("shuju.txt", ios::in);char data[str_len];Trie t1;while(ifs.getline(data, str_len, '\n') && data[0] != '\0'){t1.insert_trie(data, 1);}while(ifs>>data && data[0] != '\0'){int count = t1.find_trie(data);cout<<count<<endl;}return 0;}