Educational Codeforces Round 53 (Rated for Div. 2)G. Yet Another LCP Problem

标签：dep   sub   int   lse   ast   eps   names   scan   inline   

题意:给串s,每次询问k个数a,l个数b,问a和b作为后缀的lcp的综合
题解:和bzoj3879类似,反向sam日神仙...lcp就是fail树上的lca.把点抠出来建虚树,然后在上面dp即可.(感觉之前写的svt什么玩意)

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000009
#define ld long double
//#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=1000000+10,inf=0x3f3f3f3f;

char s[N];
struct SAM{
    int last,cnt;
    int ch[N<<1][26],fa[N<<1],l[N<<1];
    int pos[N<<1],dep[N<<1],f[N<<1][20],dfn[N<<1],res;
    vi v[N<<1];
    void ins(int c){
        int p=last,np=++cnt;last=np;l[np]=l[p]+1;
        for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
        if(!p)fa[np]=1;
        else
        {
            int q=ch[p][c];
            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][c]==q;p=fa[p])ch[p][c]=nq;
            }
        }
    }
    void build()
    {
        last=cnt=1;
        int len=strlen(s+1);
        reverse(s+1,s+1+len);
        for(int i=1;s[i];i++)ins(s[i]-'a'),pos[i]=last;
        for(int i=1;i<=cnt;i++)v[fa[i]].pb(i);
        dfs(1);
        for(int i=1;i<20;i++)for(int j=1;j<=cnt;j++)
            f[j][i]=f[f[j][i-1]][i-1];
    }
    void dfs(int u)
    {
        f[u][0]=fa[u];dfn[u]=++res;
        for(int x:v[u])
        {
//            printf("%d %d\n",u,x);
            dep[x]=dep[u]+1;
            dfs(x);
        }
    }
    int lca(int x,int y)
    {
        if(dep[x]>dep[y])swap(x,y);
        for(int i=19;~i;i--)if(((dep[y]-dep[x])>>i)&1)y=f[y][i];
        if(x==y)return x;
        for(int i=19;~i;i--)if(f[x][i]!=f[y][i])
            x=f[x][i],y=f[y][i];
        return f[x][0];
    }
}sam;
vi v[N*2],in;
void add1(int a,int b){v[a].pb(b),in.pb(a);in.pb(b);}
int st[N*2],top,a[N*2];
ll dp[N*2][2],ans;
void ins(int x)
{
    if(!top){st[++top]=x;return ;}
    int lc=sam.lca(st[top],x);
    while(top>1&&sam.dep[st[top-1]]>sam.dep[lc])
        add1(st[top-1],st[top]),top--;
    if(top>=1&&sam.dep[st[top]]>sam.dep[lc])
        add1(lc,st[top]),top--;
    if(!top||sam.dep[st[top]]<sam.dep[lc])st[++top]=lc;
    st[++top]=x;
}
bool cmp(int a,int b){return sam.dfn[a]<sam.dfn[b];}
void dfs(int u)
{
    ans+=1ll*sam.l[u]*dp[u][0]*dp[u][1];
    for(int x:v[u])
    {
        dfs(x);
        ans+=1ll*sam.l[u]*dp[u][0]*dp[x][1];
        ans+=1ll*sam.l[u]*dp[x][0]*dp[u][1];
        dp[u][0]+=dp[x][0],dp[u][1]+=dp[x][1];
    }
}
int main()
{
    int n,q;scanf("%d%d%s",&n,&q,s+1);
    sam.build();
    while(q--)
    {
        int k,l;scanf("%d%d",&k,&l);
        for(int i=0;i<k;i++)
        {
            scanf("%d",&a[i]),a[i]=sam.pos[n+1-a[i]];
            dp[a[i]][0]++;
        }
        for(int i=0;i<l;i++)
        {
            scanf("%d",&a[k+i]),a[k+i]=sam.pos[n+1-a[k+i]];
            dp[a[k+i]][1]++;
        }
        sort(a,a+k+l,cmp);k+=l;k=unique(a,a+k)-a;
        ans=top=0;ins(1);
        for(int i=0;i<k;i++)if(a[i]!=1)ins(a[i]);
        while(top>=2)add1(st[top-1],st[top]),top--;
        dfs(1);
        printf("%lld\n",ans);
        for(int x:in)v[x].clear(),dp[x][0]=dp[x][1]=0;
        in.clear();
    }
    return 0;
}
/********************

********************/

Educational Codeforces Round 53 (Rated for Div. 2)G. Yet Another LCP Problem

标签：dep   sub   int   lse   ast   eps   names   scan   inline   

原文地址：https://www.cnblogs.com/acjiumeng/p/10732422.html