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 read()
{
	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];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
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 addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
	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);
			while (top>1&&deep[stk[top-1]]>=deep[l]) addedge(stk[top-1],stk[top]),top--;
			if (l!=stk[top]) addedge(l,stk[top]),stk[top]=l;
			stk[++top]=a[i];newnode(a[i],1);
		}
		while (top) addedge(stk[top-1],stk[top]),top--;
	}
	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
	n=read(),m=read();
	scanf("%s",s+1);
	for (int i=1;i<=n;i++) ins(s[n-i+1]-'a');
	for (int i=2;i<=cnt;i++) addedge(fail[i],i);
	euler_tour::build();
	while (m--)
	{
		t=read();for (int i=1;i<=t;i++) a[i]=id[n-read()+1];
		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