[USACO08DEC] Secret Message Secret Message

 1 题目描述
 2 Bessie is leading the cows in an attempt to escape! To do this, the cows are sending secret binary messages to each other.
 3 Ever the clever counterspy, Farmer John has intercepted the first b_i (1 <= b_i <= 10,000) bits of each of M (1 <= M <= 50,000) of these secret binary messages.
 4 He has compiled a list of N (1 <= N <= 50,000) partial codewords that he thinks the cows are using. Sadly, he only knows the first c_j (1 <= c_j <= 10,000) bits of codeword j.
 5 For each codeword j, he wants to know how many of the intercepted messages match that codeword (i.e., for codeword j, how many times does a message and the codeword have the same initial bits). Your job is to compute this number.
 6 The total number of bits in the input (i.e., the sum of the b_i and the c_j) will not exceed 500,000.
 7 Memory Limit: 32MB
 8 POINTS: 270
 9 Bessie is leading the cows to escape. To contact the cows to send secret messages to each other.
10 information is binary, a total of M ( . 1 ≤M≤ 50000 ) strips. Strong anti-spyware John has the ability to intercept part of this information, Article i know the binary information before BI (L "bi≤ 10000 ) bits. He also knows that cows using N ( 1 ≤N≤ 50000 ) bar code. However, he only understand before cj Article J passwords ( 1 ≤cj≤ 10000 ) bits.
11  For each password J, he wanted to know how much information was cut and it can match. In other words, how much of this information and password have the same prefix. Of course, the prefix length must be equal to the length of the password and the smaller piece of information.
12 in the input file, the total number (i.e., bit ΣBi + ΣCi) 500000. not exceed
 13 is  input format
 14 * Line . 1 : Two integers: N and M
 15 * Lines 2 ..M + . 1 : Line I + . 1 describes intercepted code i with an integer b_i followed by b_i space-separated 0's and 1's
16 * Lines M+2..M+N+1: Line M+j+1 describes codeword j with an integer c_j followed by c_j space-separated 0's and 1's
17 输出格式
18 * Lines 1..M: Line j: The number of messages that the jth codeword could match.
19 输入输出样例
20 输入 #1 
21 4 5 
22 3 0 1 0 
23 1 . 1  
24  . 3  . 1  0  0  
25  . 3  . 1  . 1  0  
26 is  . 1  0  
27  . 1  . 1  
28  2  0  . 1  
29  . 5  0  . 1  0  0  . 1  
30  2  . 1  . 1  
31 is output # . 1  
32  . 1  
33 is  . 3  
34 is  . 1  
35  . 1  
36  2  
37 [ Description / Tips
 38  at Four messages;. Five codewords
 39 at The intercepted messages Start with 010 ,1, 100, and 110.
40 The possible codewords start with 0, 1, 01, 01001, and 11.
41 0 matches only 010: 1 match
42 1 matches 1, 100, and 110: 3 matches
43 01 matches only 010: 1 match
44 01001 matches 010: 1 match
45 11 matches 1 and 110: 2 matches

Face questions

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n,m,nw,t[500020][2],tot;
 4 int v[500020];
 5 int l,k,ans,a[500020];
 6 int b[500020];
 7 void Find()
 8 {
 9     nw=0;ans=0;
10     for(int i=1;i<=l;++i)
11     {
12         k=a[i];
13         if(!t[nw][k]){return;}
14         else{
15           nw=t[nw][k];
16           ans+=b[nw];
17         }
18     }
19     ans+=v[nw];
20     if(b[nw]) ans-=b[nw];
21 }
22 int main()
23 {
24     scanf("%d%d",&n,&m);
25     for(int i=1;i<=n;++i)
26     {
27         scanf("%d",&l);nw=0;
28         for(int j=1;j<=l;++j)
29         {
30             scanf("%d",&k);
31             if(!t[nw][k]) t[nw][k]=++tot;
32             nw=t[nw][k];v[nw]++;
33         }
34         b[nw]++;
35     }
36     while(m--)
37     {
38         scanf("%d",&l);
39         for(int i=1;i<=l;++i)
40         {
41             scanf("%d",&k);
42             a[i]=k;
43         }
44         Find();
45         printf("%d\n",ans);
46     }
47     return 0;
48 } 

View Code