Secret Message ---- (Trie树应用)

Secret Message

 

总时间限制: 

2000ms 

内存限制: 

32768kB

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

INPUT FORMAT:

* 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..M: 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

提示

INPUT DETAILS:

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

思路: Trie树的应用，注意输入的如果不是有长度一样的，需要特殊处理一下，

我是先记录一下第一个点出现的个数，然后找孩子节点，找的过程中减去孩子节点的兄弟节点的个数，

就得出了答案。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<malloc.h>
using namespace std;
typedef struct node{
	int v;
	node *next[2];
}Trie;
Trie root;
void createTrie(char *str)
{
    int len = strlen(str);
    Trie *p = &root, *q;
    for(int i=0; i<len; ++i)
    {
        int id = str[i]-'0';
        if(p->next[id] == NULL)
        {
            q = (Trie *)malloc(sizeof(root));
            q->v = 1;
            for(int j=0; j<2; ++j)
                q->next[j] = NULL;
            p->next[id] = q;
            p = p->next[id];
        }
        else
        {
            p->next[id]->v++;
            p = p->next[id];
        }
    }
}

int findTrie(char *str)
{
    int len = strlen(str);
       Trie *p = &root;
       Trie *other;
    if(len==1){
    	return 	p->next[str[0]-'0']->v;
    }
    int total =p->next[str[0]-'0']->v;
    p = p->next[str[0]-'0'];
    for(int i=1;i<len;i++){
		if(p->next[(str[i]-'0')^1]!=NULL){ // 下一个节点的兄弟节点不空，要减去
			total=total-p->next[(str[i]-'0')^1]->v;
		}
		if(i==len-1){
			break;
		}
		if(p->next[str[i]-'0']!=NULL){ //下一个节点不空，遍历
			p=p->next[str[i]-'0'];
		}
		else if(p->next[str[i]-'0']==NULL){
		    break;
		}
    }
    return total;
}

int main(){
	int n,m,i,j,k,ans,num;
	char str[50001];
	scanf("%d%d",&n,&m);
	for(i=0;i<n;i++){
		scanf("%d",&num);
		memset(str,'0',sizeof(str));
		for(j=0;j<num;j++){
			scanf("%d",&k);
			str[j]=k+'0';
		}
		str[j]='\0';
		createTrie(str);
	}
	for(i=0;i<m;i++){
		scanf("%d",&num);
	    memset(str,'0',sizeof(str));
	    for(j=0;j<num;j++){
			scanf("%d",&k);
			str[j]=k+'0';
	    }
	    str[j]='\0';
	    ans = findTrie(str);
	    printf("%d\n",ans);
	}
	return 0;
}