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2019-12-21 19:47:00
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考虑如何快速找出拥有(与它的前缀相同的)后缀的串。这个东西可以通过把字符串放到Trie里面。 分类两种情况,短串+长串长串+短串。对于短串+长串的情况,先将字符串按len排序,然后顺序将字符串倒着插入trie里面,并在trie中的尾节点记录hash。利用当前串作为查询串来找答案,这样能保证当前串一定是长串,并且找到的所有短串都有和它的前缀相同的后缀,所以再利用hash判断一下就可以确定拼出来的是不是回文串了。另外一个情况同理。 复杂度是O(26\sum len)

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 200011
#define M 1000011
#define base 233
int n;
char s[M];
int bg[N],ed[N],pos[N],c[M][27],v[M],cnt[M],len,sz=0;
ull  h[M],g[M],p[M];
il bool cmp(int x,int y){return ed[x]-bg[x]<ed[y]-bg[y];}
il int idx(char c){return c-'a';}
il void ins(int i){
    int rt=0;
    Fdr(j,ed[i],bg[i]){
        int k=idx(s[j]);
        if(!c[rt][k])c[rt][k]=++sz;
        rt=c[rt][k];
    }
    v[rt]=i;cnt[rt]++;
}
il void INS(int i){
    int rt=0;
    Fur(j,bg[i],ed[i]){
        int k=idx(s[j]);
        if(!c[rt][k])c[rt][k]=++sz;
        rt=c[rt][k];
    }
    v[rt]=i;cnt[rt]++;
}
il ull gh(int l,int r){return h[r]-h[l-1]*p[r-l+1];}
il ull gg(int l,int r){return g[l]-g[r+1]*p[r-l+1];}
il bool chk(int x,int y){
    ull a=gh(bg[x],ed[x])*p[ed[y]-bg[y]+1]+gh(bg[y],ed[y]);
    ull b=gg(bg[y],ed[y])*p[ed[x]-bg[x]+1]+gg(bg[x],ed[x]);
    return a==b;
}
int main(){
    in>>n;
    Fur(i,1,n){
        bg[i]=ed[i]=ed[i-1]+1;
        in>>(s+bg[i]);
        ed[i]+=strlen(s+bg[i])-1;
    }
    len=ed[n];
    p[0]=1;
    Fur(i,1,len){
        h[i]=h[i-1]*base+s[i];
        p[i]=p[i-1]*base;
    }
    Fdr(i,len,1)g[i]=g[i+1]*base+s[i];
    Fur(i,1,n)pos[i]=i;
    sort(pos+1,pos+n+1,cmp);
    ll ans=0;
    Fur(t,1,n){
        int i=pos[t],rt=0;
        Fur(j,bg[i],ed[i]){
            int k=idx(s[j]);
            if(!c[rt][k])break;
            if(v[rt]&&chk(i,v[rt]))ans+=cnt[rt];
            rt=c[rt][k];
        }
        if(v[rt]&&chk(i,v[rt]))ans+=cnt[rt];
        ins(i);
    }
    clr(c,0);clr(cnt,0);clr(v,0);sz=0;
    Fur(t,1,n){
        int i=pos[t],rt=0;
        Fdr(j,ed[i],bg[i]){
            int k=idx(s[j]);
            if(!c[rt][k])break;
            if(v[rt]&&chk(v[rt],i))ans+=cnt[rt];
            rt=c[rt][k];
        }
        if(v[rt]&&chk(v[rt],i))ans+=cnt[rt];
        INS(i);
    }
    out<<ans<<ln;
}
51nod 2372 回文串统计
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