2593: Secret Message(字典树)

2593: Secret Message 

时间限制(普通/Java):1000MS/10000MS     内存限制:65536KByte
总提交: 10            测试通过:3

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.

输入

 

* 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

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

样例输入

 

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

样例输出

 

1
3
1
1
2

题目来源

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <unordered_map>
#define ull unsigned long long   //溢出部分自动取余
using namespace std;

int n,m,tot;   //tot记录下标的
const int maxn=5e5+5;
int trie[maxn<<2][5],sum[maxn<<2],num[maxn<<2];

void Insert(int t){
    int root=0;
    for(int i=1,d;i<=t;i++){
        cin>>d;
        if(!trie[root][d]) trie[root][d]=++tot;
        sum[root]++;   //用来记录有几个
        root=trie[root][d];
    }
    num[root]++;  //以这个根结尾的字符串有多少个

}

int Search(int t){
    int root=0,res=0;
    for(int i=1,d;i<=t;i++){
        cin>>d;
        if(!trie[root][d]){
            for(int j=i+1,d;j<=t;j++) cin>>d;
            return res;
        }
        root=trie[root][d];
        res+=num[root];
    }
    return res+sum[root];
}

int main(){
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1,d;i<=n;i++){
        cin>>d;
        Insert(d);
    }
    for(int i=1,d;i<=m;i++){
        cin>>d;
        cout << Search(d) << endl;
    }
    return 0;
}

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <unordered_map>
#define ull unsigned long long   //溢出部分自动取余
using namespace std;

int n,m,tot;   //tot记录下标的
const int maxn=5e5+5;
int trie[maxn<<2][5],sum[maxn<<2],num[maxn<<2];

void Insert(int t){
    int root=0;
    for(int i=1,d;i<=t;i++){
        cin>>d;
        if(!trie[root][d]) trie[root][d]=++tot;
        sum[trie[root][d]]++;   //用来记录有几个
        root=trie[root][d];
    }
    num[root]++;  //以这个根结尾的字符串有多少个

}

int Search(int t){
    int root=0,res=0;
    for(int i=1,d;i<=t;i++){
        cin>>d;
        if(!trie[root][d]){
            for(int j=i+1,d;j<=t;j++) cin>>d;
            return res;
        }
        if(i!=t)   //不然的话会多加
        res+=num[trie[root][d]];
        root=trie[root][d];
    }
    return res+sum[root];   //当上面的字符串比下面的字符串长时
}

int main(){
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1,d;i<=n;i++){
        cin>>d;
        Insert(d);
    }
    for(int i=1,d;i<=m;i++){
        cin>>d;
        cout << Search(d) << endl;
    }
    return 0;
}