洛谷$P4503\ [CTSC2014]$企鹅$QQ$ 哈希

正解:哈希

解题报告:

传送门$QwQ$

直接$O(len)$枚举哪一位,然后把这一位删了重新拼接起来,存桶里查下就成

$over$?

主要的难点大概在卡哈希+卡常$QAQ$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define mp make_pair
#define ri register int
#define rc register char
#define rb register bool
#define ull unsigned long long
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)

const int N=200+10,M=30000+10,bas1=19260817,bas2=19491001;
const ull mod1=20190816170251;
int n,l,as;
char str[N];
ull poww[N],hsh1[M][N],hsh2[M][N],h1[M][N],h2[M][N],p[N];
pair<ull,ull>P[M];

il int read()
{
    rc ch=gc;ri x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:-x;
}
il void solv(ri x)
{
    rp(i,1,n)
    {ri tmp=(hsh1[i][x-1]*poww[l-x+1]%mod1+hsh2[i][x+1])%mod1,t=h1[i][x-1]*1007+h2[i][x+1]*197;P[i]=mp(tmp,t);}
    sort(P+1,P+1+n);for(ri j=2,s=1;j<=n;++j)if(P[j]==P[j-1])as+=s++;else s=1;
    //printf("%d\n",as);
}

int main()
{
    //freopen("4503.in","r",stdin);freopen("4503.out","w",stdout);
    n=read();l=read();read();poww[0]=1;rp(i,1,l)poww[i]=poww[i-1]*bas1%mod1,p[i]=p[i-1]*bas2;
    rp(i,1,n)
    {
        scanf("%s",str+1);
        rp(j,1,l)hsh1[i][j]=(hsh1[i][j-1]*bas1%mod1+str[j])%mod1,h1[i][j]=h1[i][j-1]*bas2+str[j];
        my(j,l,1)hsh2[i][j]=(hsh2[i][j+1]*bas1%mod1+str[j])%mod1,h2[i][j]=h2[i][j+1]*bas2+str[j];
    }
    rp(i,1,l)solv(i);;printf("%d\n",as);
    return 0;
}