题意：给n个字符串，给m个询问，每个询问给k个条形码。每个条形码由8个小码组成，每个小码有相应的宽度，已知一个条形码的宽度只有2种，宽的表示1，窄的表示0。并且宽的宽度是窄的宽度的2倍。由于扫描的时候有误差，每个小码的宽度为一个浮点型数据，保证每个数据的误差在5%内。所以一个条形码可以对应一个ASCC码，表示一个小写字母。k个条形码表示一个字符串s，每个询问表示给定的m个字符串中以s为前缀的字符串个数。
析：很容易看起来是Tire树， 不知道能不能暴过去，我觉得差不多可以，主要是在处理浮点数上，这个很简单，既然误差这么小，那么我们就可以取最大值和最小值然后除2，小的为0， 大的为1。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 30 * 10000 + 10;
const int mod = 10;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r > 0 && r <= n && c > 0 && c <= m;
}
struct Trie{
int ch[maxn][26];
int val[maxn];
int sz;
void init(){
memset(ch[0], 0, sizeof ch[0]);
sz = 1;
}
int idx(char c){ return c - ‘a‘; }
void insert(char *s){
int u = 0;
for(int i = 0; s[i]; ++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];
++val[u];
}
}
int query(char *s){
int u = 0;
for(int i = 0; s[i]; ++i){
int c = idx(s[i]);
if(!ch[u][c]) return 0;
u = ch[u][c];
}
return val[u];
}
};
char s[50];
Trie trie;
double a[10];
int main(){
while(scanf("%d %d", &n, &m) == 2){
trie.init();
for(int i = 0; i < n; ++i){
scanf("%s", s);
trie.insert(s);
}
LL ans = 0;
while(m--){
int k;
scanf("%d", &k);
for(int i = 0; i < k; ++i){
double mmin = inf, mmax = -1;
for(int j = 0; j < 8; ++j){
scanf("%lf", a+j);
mmin = min(mmin, a[j]);
mmax = max(mmax, a[j]);
}
double ave = (mmin + mmax) / 2.0;
int t = 0;
for(int j = 0; j < 8; ++j)
if(a[j] < ave) t = t * 2;
else t = t * 2 + 1;
s[i] = t;
}
s[k] = 0;
ans += trie.query(s);
}
printf("%I64d\n", ans);
}
return 0;
}
HDU 3724 Encoded Barcodes (Trie)
原文：http://www.cnblogs.com/dwtfukgv/p/7257725.html