[USACO08DEC] 秘密消息Secret Message

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

输入样例#1：

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

输出样例#1：

1

3

1

1

2

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

首先,我们要知道,如果是在后面m个询问中寻找前n个数字组有多少个是第\(M_i\)个数字组的前缀,那这道题就是一道模板题,详情见前缀统计

但是这道题,除此之外,因外根据题目描述,在后m个询问中,如果\(m_i\)的前缀与\(n_i\)的前缀匹配,也可以算一种方案,所以我们还要在插入时我们除了要记录该节点是多少个数字组的末尾节点cnt[],还要记录它的儿子个数size[]

更新方式:在遍历的过程中,如果不是末尾节点,就加上cnt[],否则加上size[],因为这时我们的密码已经遍历完了,但是此时这个节点在前面的n个数字组中并没有统计完,这依然可以算一种方案(或多种),所以还要加上当前节点的儿子个数size[],代表还可以匹配size[]个数字组

#include<bits/stdc++.h>
#define in(i) (i=read())
using namespace std;
int read() {
    int ans=0,f=1; char i=getchar();
    while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
    while(i>='0' && i<='9') {ans=(ans<<1)+(ans<<3)+i-'0'; i=getchar();}
    return ans*f;
}
int n,m,tot;
int trie[500010][2],a[5000010],cnt[500010],size[500010];
void insert(int k) {
    int p=0;
    for(int i=1;i<=k;i++) {
        if(!trie[p][a[i]]) trie[p][a[i]]=++tot;
        p=trie[p][a[i]];
        size[p]++;
    }
    cnt[p]++;
}
int find(int k) {
    int ans=0,p=0;
    for(int i=1;i<=k;i++) {
        p=trie[p][a[i]];
        if(!p) return ans;
        if(i!=k) ans+=cnt[p];
        else ans+=size[p];
    }
    return ans;
}
int main()
{
    int k; in(n); in(m);
    for(int i=1;i<=n;i++) {
        in(k); for(int j=1;j<=k;j++) in(a[j]);
        insert(k);
    }
    memset(a,0,sizeof(a));
    for(int i=1;i<=m;i++) {
        in(k); for(int j=1;j<=k;j++) in(a[j]);
        printf("%d\n",find(k));
    }
}

博主蒟蒻,随意转载.但必须附上原文链接

http://www.cnblogs.com/real-l/