题目描述
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
POINTS: 270
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．
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．
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．
在输入文件中，位的总数（即∑Bi+∑Ci）不会超过500000.
输入输出格式
输入格式：* 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
输出格式：* Lines 1..M: Line j: The number of messages that the jth codeword could match.
输入输出样例
说明
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
题解
这种很多个串要匹配前缀什么的，就上AC自动姬就好了。
在建Trie树的同时记录经过每个点的串个数($pas$)，和在每个点结尾的串个数($tag$)。
匹配的时候，在匹配路上加上途径的$tag$，在匹配结尾加上结尾点的$pas$。
然后如果在匹配中途就无路可走了的话，就到此为止，也不要加结尾点的$pas$。
注意一下边界就好啦。
1 /* 2 qwerta 3 P2922 [USACO08DEC]秘密消息Secret Message 4 Accepted 5 100 6 代码 C++，1.02KB 7 提交时间 2018-10-16 08:14:21 8 耗时/内存 9 709ms, 3964KB 10 */ 11 #include<iostream> 12 #include<cstdio> 13 using namespace std; 14 struct emm{ 15 int nxt[2],tag,pas; 16 }a[500003]; 17 int main() 18 { 19 //freopen("a.in","r",stdin); 20 int n,m; 21 scanf("%d%d",&n,&m); 22 int cnt=0; 23 for(int i=1;i<=n;++i) 24 { 25 int c; 26 scanf("%d",&c); 27 int k=0; 28 //建Trie树 29 for(int j=1;j<=c;++j) 30 { 31 int x; 32 scanf("%d",&x); 33 if(!a[k].nxt[x]) 34 a[k].nxt[x]=++cnt; 35 a[k].pas++; 36 k=a[k].nxt[x]; 37 } 38 //cout<<k<<endl; 39 a[k].tag++; 40 } 41 for(int i=1;i<=m;++i) 42 { 43 int c; 44 scanf("%d",&c); 45 int k=0,ans=0; 46 int flag=0; 47 for(int j=1;j<=c;++j) 48 { 49 int x; 50 scanf("%d",&x); 51 //cout<<k<<" "<<x<<" "<<a[k].nxt[x]<<endl; 52 if(!a[k].nxt[x])flag++;//如果无路可走就记个flag 53 if(!flag)//如果没有过flag就走 54 k=a[k].nxt[x], 55 ans+=a[k].tag;//加上途径的tag 56 } 57 if(!flag)//如果一路走道了底 58 cout<<ans+a[k].pas<<endl;//就加上结尾的pas 59 else 60 cout<<ans<<endl;//否则只输出途径的tag 61 } 62 return 0; 63 }