Problem
给N个字符串Si,进行M次询问,每次给定一个字符串S,问是否存在一个Si使得S与Si长度相等,同时有且只有一个字符不同。
Limits
TimeLimit(ms):3000
MemoryLimit(MB):256
N,M∈[0,3×105]
所有串总长不超过6×105
Look up Original Problem From here
Solution
字典树才是正解,但用hash会很方便。
双hash,seed=1e4级别的素数,mod1=1e9+7,mod2=1e9+9
Complexity
TimeComplexity:O(|S|all×log2N)
MemoryComplexity:O(|S|max+N)
My Code
//Hello. I‘m Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
//typedef unsigned long long ull;
#define peter cout<<"i am peter"<<endl
#define input freopen("data.txt","r",stdin)
#define randin srand((unsigned int)time(NULL))
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define pb(a) push_back(a)
#define clr(a) memset(a,0,sizeof(a))
#define clr_minus1(a) memset(a,-1,sizeof(a))
#define clr_INT(a) memset(a,INT,sizeof(a))
#define clr_true(a) memset(a,true,sizeof(a))
#define clr_false(a) memset(a,false,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define pi acos(-1.0)
#define eps 1e-9
#define MOD 1000000007
#define MAXN 1001000
#define N
#define M
ll idx(char c){
return c-‘a‘+1;
}
ll seed1[MAXN],seed2[MAXN];
const ll seed=49681;
const ll mod1=1e9+7;
const ll mod2=1e9+9;
int n,m,lens;
map<pair<ll,ll>,bool>here;
char s[MAXN];
int main(){
seed1[0]=seed2[0]=1;
rep(i,1,MAXN){
seed1[i]=seed1[i-1]*seed%mod1;
seed2[i]=seed2[i-1]*seed%mod2;
}
scanf("%d %d",&n,&m);
repin(num,1,n){
scanf("%s",s);
lens=len(s);
ll v1=0,v2=0;
rep(i,0,lens){
v1=(v1*seed+idx(s[i]))%mod1;
v2=(v2*seed+idx(s[i]))%mod2;
}
pair<ll,ll>p=make_pair(v1,v2);
here[p]=true;
}
repin(num,1,m){
scanf("%s",s);
lens=len(s);
ll v1=0,v2=0;
rep(i,0,lens){
v1=(v1*seed+idx(s[i]))%mod1;
v2=(v2*seed+idx(s[i]))%mod2;
}
ll v1t,v2t;
bool ok=false;
rep(i,0,lens){
if(ok) break;
for(char c=‘a‘;c<=‘c‘;c++){
if(s[i]==c) continue;
v1t=v1+(idx(c)-idx(s[i]))*seed1[lens-1-i];
v1t=(v1t%mod1+mod1)%mod1;
v2t=v2+(idx(c)-idx(s[i]))*seed2[lens-1-i];
v2t=(v2t%mod2+mod2)%mod2;
pair<ll,ll>p=make_pair(v1t,v2t);
if(here.find(p)!=here.end()){
ok=true;
break;
}
}
}
printf("%s\n",ok?"YES":"NO");
}
}
时间: 2024-10-24 16:01:25