bzoj 1590 luogu2292 [USACO08DEC]秘密消息Secret Message（题解）

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.
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.

样例数据

input

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

output

1 3 1 1 2

数据规模与约定

时间限制：1s
空间限制：32MB

思路：
这是一道非常有助于你理解Trie树的题，首先建立一课Trie树，用size记录儿子节点，用sum数组记录末尾节点，最后遍历Trie树，非末尾节点加上size，末尾节点加上sum

代码：

#include<bits/stdc++.h>using namespace std;const int MAXN=500010;int n,m;int a[MAXN];int k;int tree[MAXN][2];int tot;int sum[MAXN],size[MAXN];int read(){    int sum=0,flag=1;    char c;    for(;c<'0'||c>'9';c=getchar())if(c=='-') flag=-1;    for(;c>='0'&&c<='9';c=getchar())sum=(sum<<1)+(sum<<3)+c-'0';    return sum*flag;}void add(int k){    int p=0;    for(int i=1;i<=k;p=tree[p][a[i]],++i)    {        if(!tree[p][a[i]]) tree[p][a[i]]=++tot;        size[p]++;    }    sum[p]++;    size[p]++;}void work(int k){    int ans=0,p=0;    for(int i=1;i<=k;p=tree[p][a[i]],++i)    {        if(!tree[p][a[i]]) break;        if(i!=k) ans+=sum[tree[p][a[i]]];else ans+=size[tree[p][a[i]]];    }    printf("%d\n",ans);}int main(){    n=read();m=read();tot=0;    for(int i=1;i<=n;++i)    {        k=read();        for(int j=1;j<=k;++j) a[j]=read();        add(k);    }    for(int i=1;i<=m;++i)    {        k=read();        for(int j=1;j<=k;++j) a[j]=read();        work(k);    }    return 0;}