题目描述
题解
我们可以先对trie树建出广义SAM，然后维护一下right集合大小(注意right集合在广义SAM上的维护方式)。
然后把匹配穿往广义SAM上匹配，假设现在匹配到了x节点，那么x的所有祖先后可以被匹配上，那么一个节点的贡献即为r[x]*(l[x]-l[fa[x]])。
维护这玩意的和就好了，最下面的节点特判一下。
代码
#include<iostream> #include<cstdio> #include<cstring> #define N 1600002 #define M 8000002 using namespace std; typedef long long ll; char c[1],s[M]; int cnt,n,father,pa[N],len,tong[N],rnk[N]; ll sum[N],ans; int l[N],ch[N][26],fa[N],r[N]; inline int rd(){ int x=0;char c=getchar();bool f=0; while(!isdigit(c)){if(c=='-')f=1;c=getchar();} while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();} return f?-x:x; } inline int ins(int last,int x){ int p=last; if(ch[p][x]){ int q=ch[p][x]; if(l[p]+1==l[q]){r[q]++;return q;} else{ int nq=++cnt;l[nq]=l[p]+1;r[nq]=1;//care!!!!!!!!!!!!!!! memcpy(ch[nq],ch[q],sizeof(ch[q])); fa[nq]=fa[q];fa[q]=nq; for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq; return nq; } } else{ int np=++cnt;l[np]=l[p]+1;r[np]=1; for(;p&&!ch[p][x];p=fa[p])ch[p][x]=np; if(!p)fa[np]=1; else{ int q=ch[p][x]; if(l[p]+1==l[q])fa[np]=q; else{ int nq=++cnt;l[nq]=l[p]+1; memcpy(ch[nq],ch[q],sizeof(ch[q])); fa[nq]=fa[q];fa[q]=fa[np]=nq; for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq; } } return np; } } int main(){ n=rd();cnt=1;pa[1]=1; for(int i=2;i<=n;++i){ father=rd();scanf("%s",c); pa[i]=ins(pa[father],c[0]-'a'); } scanf("%s",s+1);len=strlen(s+1); for(int i=1;i<=cnt;++i)tong[l[i]]++; for(int i=1;i<=n;++i)tong[i]+=tong[i-1]; for(int i=1;i<=cnt;++i)rnk[tong[l[i]]--]=i; for(int i=cnt;i>=1;--i){int x=rnk[i];r[fa[x]]+=r[x];} r[1]=0; for(int i=1;i<=cnt;++i){int x=rnk[i];sum[x]=sum[fa[x]]+1ll*(l[x]-l[fa[x]])*r[x];} int now=1,le=0; for(int i=1;i<=len;++i){ if(ch[now][s[i]-'a'])le++,now=ch[now][s[i]-'a']; else{ for(;now&&!ch[now][s[i]-'a'];now=fa[now]); if(now)le=l[now]+1,now=ch[now][s[i]-'a']; else le=0,now=1; } if(now!=1)ans+=sum[fa[now]]+1ll*(le-l[fa[now]])*r[now]; } cout<<ans; return 0; }
#include<iostream> #include<cstdio> #include<cstring> #define N 1600002 #define M 8000002 using namespace std; typedef long long ll; char c[1],s[M]; int cnt,n,father,pa[N],len,tong[N],rnk[N]; ll sum[N],ans; int l[N],ch[N][26],fa[N],r[N]; inline int rd(){ int x=0;char c=getchar();bool f=0; while(!isdigit(c)){if(c=='-')f=1;c=getchar();} while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();} return f?-x:x; } inline int ins(int last,int x){ int p=last; if(ch[p][x]){ int q=ch[p][x]; if(l[p]+1==l[q]){r[q]++;return q;} else{ int nq=++cnt;l[nq]=l[p]+1;r[nq]=1;//care!!!!!!!!!!!!!!! memcpy(ch[nq],ch[q],sizeof(ch[q])); fa[nq]=fa[q];fa[q]=nq; for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq; return nq; } } else{ int np=++cnt;l[np]=l[p]+1;r[np]=1; for(;p&&!ch[p][x];p=fa[p])ch[p][x]=np; if(!p)fa[np]=1; else{ int q=ch[p][x]; if(l[p]+1==l[q])fa[np]=q; else{ int nq=++cnt;l[nq]=l[p]+1; memcpy(ch[nq],ch[q],sizeof(ch[q])); fa[nq]=fa[q];fa[q]=fa[np]=nq; for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq; } } return np; } } int main(){ n=rd();cnt=1;pa[1]=1; for(int i=2;i<=n;++i){ father=rd();scanf("%s",c); pa[i]=ins(pa[father],c[0]-'a'); } scanf("%s",s+1);len=strlen(s+1); for(int i=1;i<=cnt;++i)tong[l[i]]++; for(int i=1;i<=n;++i)tong[i]+=tong[i-1]; for(int i=1;i<=cnt;++i)rnk[tong[l[i]]--]=i; for(int i=cnt;i>=1;--i){int x=rnk[i];r[fa[x]]+=r[x];} r[1]=0; for(int i=1;i<=cnt;++i){int x=rnk[i];sum[x]=sum[fa[x]]+1ll*(l[x]-l[fa[x]])*r[x];} int now=1,le=0; for(int i=1;i<=len;++i){ if(ch[now][s[i]-'a'])le++,now=ch[now][s[i]-'a']; else{ for(;now&&!ch[now][s[i]-'a'];now=fa[now]); if(now)le=l[now]+1,now=ch[now][s[i]-'a']; else le=0,now=1; } if(now!=1)ans+=sum[fa[now]]+1ll*(le-l[fa[now]])*r[now]; } cout<<ans; return 0; }