• [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
    ",ans);
    46     }
    47     return 0;
    48 } 
    View Code
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  • 原文地址:https://www.cnblogs.com/adelalove/p/11778657.html
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