題目描述

Bessie is leading the cows in an attempt to escape! To do this, the cows are sending secret binary messages to each other.

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.

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.

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.

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.

Memory Limit: 32MB

輸入輸出格式

輸入格式：

* Line 1: Two integers: M and N

* 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

* 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

輸出格式：

* Lines 1..M: Line j: The number of messages that the jth codeword could match.

輸入輸出樣例

4 5 
3 0 1 0 
1 1 
3 1 0 0 
3 1 1 0 
1 0 
1 1 
2 0 1 
5 0 1 0 0 1 
2 1 1 

1 
3 
1 
1 
2 

說明

Four messages; five codewords.

The intercepted messages start with 010, 1, 100, and 110.

The possible codewords start with 0, 1, 01, 01001, and 11.

0 matches only 010: 1 match

1 matches 1, 100, and 110: 3 matches

01 matches only 010: 1 match

01001 matches 010: 1 match

11 matches 1 and 110: 2 matches

第一道英文題，有點小激動……

由題意知我們可以把0 1看成字符串，設sum表示經過的已知字符個數，end表示以該節點結束的字符……

然後第一種：詢問到盡頭，那就是答案-end+sum（想想，為什麽？），

不然就是答案。

AC代碼如下（看過其他題解，不一定最快，但是最簡單的）

#include<cstdio>
using namespace std;
const int N=500000+5;
int ch[N][2],sum[N],end[N];
int x,l,n,m,u,tot,ans;
bool ff;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&l);
        u=0;
        for(int j=1;j<=l;j++)
        {
            scanf("%d",&x);
            if(!ch[u][x]) ch[u][x]=++tot;
            u=ch[u][x];
            sum[u]++;
        }
        end[u]++;
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d",&l);
        ans=u=0;ff=1;
        for(int j=1;j<=l;j++)
        {
            scanf("%d",&x);
            if(ff&&ch[u][x]) u=ch[u][x];
            else ff=0;
            if(ff) ans+=end[u];
        }
        if(ff) printf("%d\n",ans-end[u]+sum[u]);
        else printf("%d\n",ans);
    }
    return 0;
}

P2922 [USACO08DEC]秘密消息Secret Message