[USACO08DEC]祕密消息Secret Message

 1 題目描述
 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 貝茜正在領導奶牛們逃跑．爲了聯絡，奶牛們互相發送祕密信息．
10 信息是二進制的，共有M(1≤M≤50000)條．反間諜能力很強的約翰已經部分攔截了這些信息，知道了第i條二進制信息的前bi(l《bi≤10000)位．他同時知道，奶牛使用N(1≤N≤50000)條密碼．可是，他僅僅瞭解第J條密碼的前cj(1≤cj≤10000)位．
11 對於每條密碼J，他想知道有多少截得的信息可以和它匹配．也就是說，有多少信息和這條密碼有着相同的前綴．固然，這個前綴長度必須等於密碼和那條信息長度的較小者．
12 在輸入文件中，位的總數（即∑Bi+∑Ci）不會超過500000.
13 輸入格式
14 * Line 1: Two integers: M and N
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 輸出格式
18 * Lines 1..M: Line j: The number of messages that the jth codeword could match.
19 輸入輸出樣例
20 輸入 #1 
21 4 5 
22 3 0 1 0 
23 1 1 
24 3 1 0 0 
25 3 1 1 0 
26 1 0 
27 1 1 
28 2 0 1 
29 5 0 1 0 0 1 
30 2 1 1 
31 輸出 #1 
32 1 
33 3 
34 1 
35 1 
36 2 
37 說明/提示
38 Four messages; five codewords.
39 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

題面

 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 } 

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