POJ3349 Snowflake Snow Snowflakes 哈希

注意到这个在哈希时要考虑循环同构。。。所以哈希函数是$sum a_i+prod a_i$

如果发现有哈希值相等的了就$ck$一下。

最后$ck$时可以用最小表示法的知识提高效率，，，先咕着。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#define ll long long
#define R register int
using namespace std;
namespace Fread {
    static char B[1<<15],*S=B,*D=B;
    #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() {
        R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
        do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
    }
}using Fread::g;
const int N=100010,p=99991;
int n;
struct node {
    int fir[N],nxt[N],s[N][6],cnt; 
    inline void clear() {
        memset(fir,0,sizeof(fir)),memset(nxt,0,sizeof(nxt));
        memset(s,0,sizeof(s)),cnt=0;
    }
    inline int hsh(int* d) { R sum=0,mul=1;
        for(R i=0;i<6;++i) sum=(sum+d[i])%p,mul=(ll)mul*d[i]%p; return (sum+mul)%p;
    }
    inline bool ck(int* a,int* b) {
        for(R i=0;i<6;++i) for(R j=0;j<6;++j) {
            register bool flg=1;
            for(R k=0;k<6;++k) if(a[(i+k)%6]!=b[(j+k)%6]) flg=0;
            if(flg) return 1; flg=1;
            for(R k=0;k<6;++k) if(a[(i+k)%6]!=b[(j-k+6)%6]) flg=0;
            if(flg) return 1; 
        } return 0;
    }
    inline bool ins(int* d) {
        R st=hsh(d); for(R i=fir[st];i;i=nxt[i]) if(ck(s[i],d)) return 1;
        ++cnt; memcpy(s[cnt],d,sizeof(int)*6); nxt[cnt]=fir[st],fir[st]=cnt; return 0;
    }
}HSH;
signed main() {
#ifdef JACK
    freopen("NOIPAK++.in","r",stdin);
#endif
    n=g(); for(R i=1;i<=n;++i) { R x[7];
        for(R j=0;j<6;++j) x[j]=g();        
        if(HSH.ins(x)) { printf("Twin snowflakes found.
"); return 0;}
    } printf("No two snowflakes are alike.
");
}

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2019.06.10