【题解】P4503 [CTSC2014]企鹅QQ(哈希)

【题解】P4503 [CTSC2014]企鹅QQ(哈希)

考虑这样一种做法，将每个字符串的删去某个字符的新字符串的哈希值存下来，然后最后\(sort\)一遍双指针统计每个值相同的数的个数\(x\)，这个\(x\)对答案的贡献是\({x \choose 2}\)

//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;  typedef long long ll;
inline int qr(){
      register int ret=0,f=0;
      register char c=getchar();
      while(c<48||c>57)f|=c==45,c=getchar();
      while(c>=48&&c<=57) ret=ret*10+c-48,c=getchar();
      return f?-ret:ret;
}
const int maxs=2e2+5;
int n,len,S,cnt;
char c[maxs];
int ha1[maxs],ha2[maxs];
int s1[maxs],s2[maxs];
int base1[maxs],base2[maxs];
const int mod1=998244353,mod2=993244853;
const int seed1=233,seed2=666;
pair<int,int> data[200*30001];

inline void insert(const char*c){
      for(int t=1;t<=len;++t) ha1[t]=(1ll*ha1[t-1]*seed1+c[t])%mod1,ha2[t]=(1ll*ha2[t-1]*seed2+c[t])%mod2;
      for(int t=len;t;--t) s1[t]=(1ll*s1[t+1]*seed1+c[t])%mod1,s2[t]=(1ll*s2[t+1]*seed2+c[t])%mod2;
      for(int t=1;t<=len;++t) data[++cnt]={(0ll+ha1[t-1]+1ll*base1[t]*s1[t+1]%mod1)%mod1,(0ll+ha2[t-1]+1ll*base2[t]*s2[t+1]%mod2)%mod2};
}

int main(){
      base1[0]=base2[0]=1;
      n=qr(); len=qr(); S=qr();
      for(int t=1;t<=len;++t) base1[t]=1ll*base1[t-1]*seed1%mod1,base2[t]=1ll*base2[t-1]*seed2%mod2;
      for(int t=1;t<=n;++t) scanf("%s",c+1),insert(c);//cout<<(c+1)<<endl;
      sort(data+1,data+cnt+1);
      ll ans=0;
      for(int t=1,r=1;t<=cnt;t=++r){
        while(data[t]==data[r+1]) ++r;
        if(r>t) ans=ans+((r-t+1LL)*(r-t)>>1);
      }
      printf("%lld\n",ans);
      return 0;
}

原文地址：https://www.cnblogs.com/winlere/p/11594969.html