[USACO08DEC]秘密消息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 贝茜正在领导奶牛们逃跑.为了联络,奶牛们互相发送秘密信息.
10 信息是二进制的,共有M(1≤M≤50000)条.反间谍能力很强的约翰已经部分拦截了这些信息,知道了第i条二进制信息的前bi(l《bi≤10000)位.他同时知道,奶牛使用N(1≤N≤50000)条密码.但是,他仅仅了解第J条密码的前cj(1≤cj≤10000)位.
11 对于每条密码J,他想知道有多少截得的信息能够和它匹配.也就是说,有多少信息和这条密码有着相同的前缀.当然,这个前缀长度必须等于密码和那条信息长度的较小者.
12 在输入文件中,位的总数(即∑Bi+∑Ci)不会超过500000.
13 输入格式
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 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 1 0 
27 1 1 
28 2 0 1 
29 5 0 1 0 0 1 
30 2 1 1 
31 输出 #1 
32 1 
33 3 
34 1 
35 1 
36 2 
37 说明/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
题面
 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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posted @ 2019-11-01 18:35  月亮茶  阅读(280)  评论(0编辑  收藏  举报