HDU 3724 Encoded Barcodes(trie樹)

Encoded Barcodes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1167    Accepted Submission(s): 402


Problem DescriptionAll 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.

 

InputThere are several test cases in the input. The first line of each case contains two integers N and M (1 <= N <= 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. 

OutputOutput one line for each test case, indicating the sum of all the query results as described above.

 

Sample Input

4 3appleappleavatarbook11 2 2 1 1 1 1 221 2 2 1 1 1 1 210.1 20.1 19.9 20.0 10.2 9.8 9.9 10.011 2 2 1 1 1 2 2

 

Sample Output

5HintThere 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. 

 

Source2010 Asia Tianjin Regional Contest 

Recommendzhouzeyong 這個題目是2010天津賽區現場賽的題目，題目比較容易，但是我wa了兩次！這次wa的原因不是程式問題，是run time error  原因是對於語言的理解上面，我定義的結構提是有建構函式的，但是我在申請的時候用的是malloc那麼這個函數是C語言裡面申請記憶體塊的一個函數，申請的就是size大小的一個記憶體塊，不會認為這段記憶體塊是分配給某個對象或者什麼的，所以不會調用建構函式，那麼就完了，肯定run time error了這個題目思路比較簡單，就是trie樹，在插入的過程中將路途中經過的所有節點的num+1，最後查詢就OK了還有那個條碼，怎樣判斷是0還是1，我是這樣的，找出最大的，和最小的，取平均，只要比平均大的就是1否則是0，這樣就OK了，然後查詢結果加起來就好了！

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int n,m;struct trie{int num;trie *next[26];trie(){num=0;memset(next,0,sizeof(next));}}re_root;int insert_trie(trie *root,char *name){if(name[0]==0){root->num++;return 0;}root->num++;int temp=int(name[0]-'a');if(root->next[temp]!=NULL)insert_trie(root->next[temp],name+1);else{root->next[temp]=new trie;insert_trie(root->next[temp],name+1);}return 0;}int query(trie *root,char *name){if(name[0]==0)return root->num;int temp=name[0]-'a';if(root->next[temp]!=NULL)return query(root->next[temp],name+1);elsereturn 0;}int main(){int i,j,k,r;int ans;char str[10000];char pattern[10000];double code[10];double Max,Min,Avg;int two;while(scanf("%d%d",&n,&m)!=EOF){re_root.num=0;memset(re_root.next,0,sizeof(re_root.next));ans=0;for(i=0;i<n;i++){scanf("%s",str);insert_trie(&re_root,str);}for(i=0;i<m;i++){scanf("%d",&k);for(j=0;j<k;j++){scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&code[0],&code[1],&code[2],&code[3],&code[4],&code[5],&code[6],&code[7]);Max=code[0];Min=code[0];for(r=1;r<8;r++){if(code[r] > Max)Max=code[r];if(code[r] < Min)Min=code[r];}two=0;Avg=(Max+Min)/2;for(r=0;r<8;r++)if(code[r] > Avg)two+=(1<<(7-r));//printf("two:%d %lf\n",two,Avg);pattern[j]=two;}pattern[k]=0;ans+=query(&re_root,pattern);}printf("%d\n",ans);}return 0;