首页 > 代码库 > UVALive 5029 Encoded Barcodes
UVALive 5029 Encoded Barcodes
Description
All the big malls need a powerful system for the products retrieval. Now you are employed design a sub-system: reading the barcodes and return the matching products.
A barcode is an optical machine-readable representation of data, which shows certain data on certain products. A barcode consists of a series of bars with different widths. In our system, the barcodes have been scanned and the widths have been recorded. Every consecutive eight bars are considered as representing the ASCII code of a character, each bar for each bit. Ideally, there should be only two kinds of widths in the eight bars, and the width of the wider bar is twice of the narrower. The wider bar indicates 1, while the narrower indicates 0. However, due to the inaccuracy of printing and scanning, there will be an error of at most 5%. That is, if the pretended exact width is x, you may get a value in the range [0.95x, 1.05x].
For example, the width sequence ``10.0 20.0 10.0 10.0 10.0 10.0 10.0 20.0" is a valid barcode of our system, and it means (01000001)2, which is (65)10 and the corresponding character is ``A". Note that ``10.5 20.1 10.1 10.2 9.9 9.7 10.0 19.9" is also a valid barcode representing the same letter.
You are given the names of all the products and many queries. Every name contains lower-case letters only, and the length is no more than 30. The queries are represented as barcodes. For each query, you should decode it to a string S, and report the amount of products whose prefix is S. For the output may be very large, you only need to output the sum of all the queries for each case.
Input
There are several test cases in the input. The first line of each case contains two integers N and M(1N
10000, 1
M
2000), indicating the number of products and queries. Then N lines follow, indicating the names of the products. Note that the names may be duplicated. Then M query blocks follow. The first line of each query block is an integer K(0 < K
30) indicating the length of the query, then K lines follow, each line contains 8 positive float numbers, indicating the barcode for each character.
You can assume that the barcodes are always valid, and always represent lower-case letters.
Output
Output one line for each test case, indicating the sum of all the query results as described above.
Explanation for the sample:
There is only one test case. The first query is ``a", and the answer is 3. The second query is ``ap", and the answer is 2. The third query is ``c", and the answer is 0. So the total sum is 3+2+0 = 5.
Sample Input
4 3 apple apple avatar book 1 1 2 2 1 1 1 1 2 2 1 2 2 1 1 1 1 2 10.1 20.1 19.9 20.0 10.2 9.8 9.9 10.0 1 1 2 2 1 1 1 2 2
Sample Output
5
/*
能力有限，只能是用暴力写的，看大神们都是用字典树写的，时间大约是100+ms
而暴力的却要800+ms，差距还是不小的。有空需要恶补一下。
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define maxn 11111
using namespace std;
typedef long long ll;
int main()
{
char s[10010][35];
double d[9];
memset(s,0,sizeof(s));
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
int ans=0;
for(int i=0;i<n;i++)
{
scanf("%s",s[i]);
}
int k;
for(int i=0;i<m;i++)
{
scanf("%d",&k);
char t[35];
memset(t,0,sizeof(t));
for(int j=0;j<k;j++)
{
double mi=0;
for(int v=0;v<=7;v++)
{
scanf("%lf",&d[v]);
mi+=d[v];
}
mi/=8.0;
//求出8个数的平均值，在平均值之上的是宽条形码，也就是1
//在平均值之下的是窄条形码，也就是0
int b=1,sum=0;
for(int v=7;v>=0;v--)
{
if(d[v]>mi)
sum+=b;
b*=2;
}
t[j]=sum;
}
//cout<<t<<endl;
for(int v=0;v<n;v++)
if(strstr(s[v],t)==s[v])//strstr函数返回子串在母串中匹配到的位置
{ //失匹返回null，等于s[v]说明是前缀
ans++;
}
}
printf("%d\n",ans);
}
return 0;
}