Watto and Mechanism（字符串hash算法）

Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with n strings. Then the mechanism should be able to process queries of the following type: “Given string s, determine if the memory of the mechanism contains string t that consists of the same number of characters as s and differs froms in exactly one position”.

Watto has already compiled the mechanism, all that’s left is to write a program for it and check it on the data consisting of n initial lines and m queries. He decided to entrust this job to you.

Input
The first line contains two non-negative numbers n and m (0 ≤ n ≤ 3·105, 0 ≤ m ≤ 3·105) — the number of the initial strings and the number of queries, respectively.

Next follow n non-empty strings that are uploaded to the memory of the mechanism.

Next follow m non-empty strings that are the queries to the mechanism.

The total length of lines in the input doesn’t exceed 6·105. Each line consists only of letters ‘a’, ‘b’, ‘c’.

Output
For each query print on a single line “YES” (without the quotes), if the memory of the mechanism contains the required string, otherwise print “NO” (without the quotes).

Sample test(s)
input
2 3
aaaaa
acacaca
aabaa
ccacacc
caaac
output
YES
NO
NO

题意：字符串只有abc三个字母，给n个字符串，m个测试字符串，对于每个测试字符串，判断n个字符串中是否有只和测试字符串差1个字母的字符串

字符串的hash，，，精髓在于把一个字符串映射成一个多项式加以比较
推荐这篇————————————————————–

做法：对于每个测试字符串，枚举一下每个位置的字符，用a,b,c替换得到一个新的hash在以前的Hash里找有没有相同的就可以了
注意set的用法

#include <iostream>#include <cmath>#include <cstring>#include <algorithm>#include <cstdio>#include <set>#define maxn 300010using namespace std;typedef long long ll;const ll mod=1e9+7;const ll key=257;int n,m;char a[1000010];ll ht[1000010];set<ll> has;void init(int n){    ht[0]=1;    for(int i=1;i<n;i++)        ht[i]=ht[i-1]*key%mod;}int gethash(char * x){    int len=strlen(x);    ll tmp=0;    for(int i=0;i<len;i++)    {        tmp=(tmp*key+x[i])%mod;    }    return tmp;}bool judge(char * x){    ll sans=gethash(x);    int len=strlen(x);    for(int i=0;i<len;i++)    {        for(ll j='a';j<='c';j++)        {            if(j==x[i])continue;            if(has.find((((j-x[i])*ht[len-i-1]+sans)%mod+mod)%mod)!=has.end())                return true;        }    }    return false;}int main(){    ll sans=0;    init(1000005);    while(scanf("%d%d",&n,&m)!=EOF)    {        has.clear();        for(int i=0;i<n;i++)        {            scanf("%s",a);            sans=gethash(a);            has.insert(sans);        }        ll t;        int flag=1;        for(int i=0;i<m;i++)        {            scanf("%s",a);            if(!judge(a))printf("NO\n");            else printf("YES\n");        }    }    return 0;}