HDU - 3724 Encoded Barcodes （字典树）

题意：给定n个字符串和m个经过处理得到的字符串，问对于m个字符串中的每个字符串，n个字符串中以该字符串为前缀的个数。
分析：
1、误差在[0.95x, 1.05x]，因此求8个数的平均数，大于平均数为1，否则为0。
2、字典树求前缀个数。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 1000000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a < b ? -1 : 1;
}
struct Trie{
    int val;
    Trie *nex[26];
    Trie(){
        val = 0;
        for(int i = 0; i < 26; ++i) nex[i] = NULL;
    }
};
void build(string x, Trie *root){
    int len = x.size();
    for(int i = 0; i < len; ++i){
        int cur = x[i] - 'a';
        if(root -> nex[cur] == NULL){
            root -> nex[cur] = new Trie();
        }
        root = root -> nex[cur];
        ++root -> val;
    }
}
int query(string x, Trie *root){
    int len = x.size();
    for(int i = 0; i < len; ++i){
        int cur = x[i] - 'a';
        if(root -> nex[cur] == NULL) return 0;
        root = root -> nex[cur];
    }
    return root -> val;
}
int POW[10];
void init(){
    POW[0] = 1;
    for(int i = 1; i <= 7; ++i){
        POW[i] = POW[i - 1] * 2;
    }
}
int main(){
    int N, M;
    init();
    while(scanf("%d%d", &N, &M) == 2){
        string s;
        Trie *root = new Trie();
        for(int i = 0; i < N; ++i){
            cin >> s;
            build(s, root);
        }
        int K;
        int ans = 0;
        while(M--){
            scanf("%d", &K);
            s = "";
            while(K--){
                double ave = 0;
                double a[10];
                for(int i = 0; i < 8; ++i){
                    scanf("%lf", &a[i]);
                    ave += a[i];
                }
                ave /= 8;
                int x = 0;
                for(int i = 0; i < 8; ++i){
                    if(a[i] > ave){
                        x += POW[7 - i] * 1;
                    }
                }
                s += x - 97 + 'a';
            }
            ans += query(s, root);
        }
        printf("%d\n", ans);
    }
    return 0;
}