BZOJ3555 [Ctsc2014]企鹅QQ[暴力+字符串hash]

菜到自闭，一道省选小水题都能给我做繁。

要求有一位不同，则对每个串每一位暴力枚举把这一位删掉，放一个分隔符，算一下hash，插表，相似的都应该会被插入同一个桶。最后把hash统计一下即可。复杂度$O(nL)$。

但是除了我以外的所有人全是把hash算好丢数组里排序，然后相似的串hash必连在一起。复杂度$O(nLlognL)$，吊打前者。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #define dbg(x) cerr << #x << " = " << x <<endl
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef double db;
 typedef pair<int,int> pii;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,):;}
 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}
 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const ull base1=;
 const int P=,MOD=;
 const int N=+,L=+;
 char s[L];
 ull H1,bek1;
 int n,l,siz,len;
 struct HASH{
     ull ha1[MOD];int Head[MOD],nxt[MOD],cnt[MOD],tot;
     inline void Insert(ull x){
         int tmp=x%MOD,id=;
         for(int j=Head[tmp];j;j=nxt[j])if(x==ha1[j]){id=j;break;}
         if(!id)ha1[++tot]=x,nxt[tot]=Head[tmp],Head[tmp]=id=tot;
         ++cnt[id];
     }
     inline int Count(){
         int ret=;
         for(register int i=;i<=tot;++i)ret+=cnt[i]*(cnt[i]-)/;
         return ret;
     }
 }H;
 
 inline void read_data(){
     for(register int i=;i<=n;++i){
         scanf("%s",s+);
         H1=;bek1=;
         for(register int j=;j<=l;++j)H1=H1*base1+(s[j]-);
         for(register int j=l;j;--j,bek1*=base1)
             H.Insert(H1-(s[j]-)*bek1);
     }
 }
 
 int main(){//freopen("test.in","r",stdin);freopen("test.ans","w",stdout);
     read(n),read(l),read(siz);
     read_data();
     return printf("%d\n",H.Count()),;
 }

hash查表

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #define dbg(x) cerr << #x << " = " << x <<endl
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef double db;
 typedef pair<int,int> pii;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,):;}
 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}
 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const ull base1=;
 const int MOD=;
 const int N=+,L=+;
 char s[L];
 ull H1,bek1,ha[MOD];
 int n,l,siz,len,ans,cnt,tot;
 inline void read_data(){
     for(register int i=;i<=n;++i){
         scanf("%s",s+);
         H1=;bek1=;
         for(register int j=;j<=l;++j)H1=H1*base1+(s[j]-);
         for(register int j=l;j;--j,bek1*=base1)
             ha[++tot]=H1-(s[j]-)*bek1;
     }
 }
 
 int main(){//freopen("test.in","r",stdin);freopen("test.ans","w",stdout);
     read(n),read(l),read(siz);
     read_data();
     sort(ha+,ha+tot+);
     ha[]=ha[]-;
     for(register int i=;i<=tot+;++i)ha[i]^ha[i-]?(ans+=(cnt*(cnt-))>>,cnt=):++cnt;
     printf("%d\n",ans);
     return ;
 }

暴力sort