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,1):0;}
template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;}
template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
template<typename T>inline T read(T&x){
    x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
    while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
}
const ull base1=131;
const int P=998244353,MOD=6000011;
const int N=30000+5,L=200+3;
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=0;
        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=0;
        for(register int i=1;i<=tot;++i)ret+=cnt[i]*(cnt[i]-1)/2;
        return ret;
    }
}H;

inline void read_data(){
    for(register int i=1;i<=n;++i){
        scanf("%s",s+1);
        H1=0;bek1=1;
        for(register int j=1;j<=l;++j)H1=H1*base1+(s[j]-47);
        for(register int j=l;j;--j,bek1*=base1)
            H.Insert(H1-(s[j]-147)*bek1);
    }
}

int main(){//freopen("test.in","r",stdin);freopen("test.ans","w",stdout);
    read(n),read(l),read(siz);
    read_data();
    return printf("%d
",H.Count()),0;
}

#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,1):0;}
template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;}
template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
template<typename T>inline T read(T&x){
    x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
    while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
}
const ull base1=131;
const int MOD=6000011;
const int N=30000+5,L=200+3;
char s[L];
ull H1,bek1,ha[MOD];
int n,l,siz,len,ans,cnt,tot;
inline void read_data(){
    for(register int i=1;i<=n;++i){
        scanf("%s",s+1);
        H1=0;bek1=1;
        for(register int j=1;j<=l;++j)H1=H1*base1+(s[j]-47);
        for(register int j=l;j;--j,bek1*=base1)
            ha[++tot]=H1-(s[j]-147)*bek1;
    }
}

int main(){//freopen("test.in","r",stdin);freopen("test.ans","w",stdout);
    read(n),read(l),read(siz);
    read_data();
    sort(ha+1,ha+tot+1);
    ha[0]=ha[1]-1;
    for(register int i=1;i<=tot+1;++i)ha[i]^ha[i-1]?(ans+=(cnt*(cnt-1))>>1,cnt=1):++cnt;
    printf("%d
",ans);
    return 0;
}