BZOJ3879 SvT（后缀树+虚树）

对反串建SAM得到后缀树，两后缀的lcp就是其在后缀树上lca的len值，于是每次询问对后缀树建出虚树并统计答案即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1000010
#define P 23333333333333333ll
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,t,a[N*3],son[N][26],fail[N],deep[N],len[N],id[N],p[N],dfn[N],cnt=1,last=1;
struct data{int to,nxt;
}edge[N];
char s[N];
void ins(int c)
{
int x=++cnt,p=last;last=x;len[x]=len[p]+1;id[len[x]]=x;
while (!son[p][c]) son[p][c]=x,p=fail[p];
if (!p) fail[x]=1;
else
{
int q=son[p][c];
if (len[p]+1==len[q]) fail[x]=q;
else
{
int y=++cnt;
len[y]=len[p]+1;
memcpy(son[y],son[q],sizeof(son[q]));
fail[y]=fail[q],fail[q]=fail[x]=y;
while (son[p][c]==q) son[p][c]=y,p=fail[p];
}
}
}
namespace euler_tour
{
int id[N<<1],LG2[N<<1],f[N<<1][22],cnt;
void dfs(int k)
{
dfn[k]=++cnt;id[cnt]=k;
for (int i=p[k];i;i=edge[i].nxt)
{
deep[edge[i].to]=deep[k]+1;
dfs(edge[i].to);
id[++cnt]=k;
}
}
void build()
{
dfs(1);
for (int i=1;i<=cnt;i++) f[i][0]=id[i];
for (int j=1;j<=21;j++)
for (int i=1;i<=cnt;i++)
if (deep[f[i][j-1]]<deep[f[min(cnt,i+(1<<j-1))][j-1]]) f[i][j]=f[i][j-1];
else f[i][j]=f[min(cnt,i+(1<<j-1))][j-1];
for (int i=2;i<=cnt;i++)
{
LG2[i]=LG2[i-1];
if ((2<<LG2[i])<=i) LG2[i]++;
}
}
int lca(int x,int y)
{
if (!x||!y) return 0;
x=dfn[x],y=dfn[y];
if (x>y) swap(x,y);
if (deep[f[x][LG2[y-x+1]]]<deep[f[y-(1<<LG2[y-x+1])+1][LG2[y-x+1]]]) return f[x][LG2[y-x+1]];
else return f[y-(1<<LG2[y-x+1])+1][LG2[y-x+1]];
}
}
using euler_tour::lca;
namespace virtual_tree
{
int p[N],size[N],stk[N],top,t;
bool flag[N];
ll ans;
struct data{int to,nxt;}edge[N];
void newnode(int k,int x){if (!flag[k]) p[k]=0,flag[k]=1,size[k]=x;}
void build(int *a,int n)
{
stk[top=1]=1;newnode(1,0);t=0;
for (int i=1;i<=n;i++)
{
int l=lca(a[i],stk[top]);newnode(l,0);
stk[++top]=a[i];newnode(a[i],1);
}
}
void work(int k)
{
flag[k]=0;
for (int i=p[k];i;i=edge[i].nxt)
{
work(edge[i].to);
ans=(ans+1ll*size[k]*size[edge[i].to]*len[k])%P;
size[k]+=size[edge[i].to];
}
}
ll calc()
{
ans=0;
work(1);
return ans;
}
}
bool cmp(const int&x,const int&y)
{
return dfn[x]<dfn[y];
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
scanf("%s",s+1);
for (int i=1;i<=n;i++) ins(s[n-i+1]-'a');
euler_tour::build();
while (m--)
{
sort(a+1,a+t+1,cmp);t=unique(a+1,a+t+1)-a-1;
virtual_tree::build(a,t);
printf(LL,virtual_tree::calc());
}
return 0;
}


posted @ 2019-05-08 17:19  Gloid  阅读(149)  评论(0编辑  收藏  举报