[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
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 }