bzoj 3626: [LNOI2014]LCA 离线+树链剖分

共5组数据，n与q的规模分别为10000,20000,30000,40000,50000。

　　真是不知道那些人是如何想到这道题正解的，首先，区间查询都可以转化为前缀和的差值（就这个我都没想到），然后前缀和就想到了离线，按照点编号加点，查询[1,a]与z的答案，可以将z至根的路径赋值为1，询问[1,a]每个点到更路径点权和，而本题最神奇的地方是可以转化为将[1,a]所有点到根路径点权加1，询问z至根的点权和。

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 51100
#define MAXE MAXN*2
#define MAXQ MAXN
#define MAXV MAXN
#define MOD 201314
#define lch (now<<1)
#define rch (now<<1^1)
int n,m;
struct sgt_node
{
        int l,r,sum,lzy;
}sgt[MAXN*4];
void up(int now)
{
        if (sgt[now].l==sgt[now].r)return;
        sgt[now].sum=(sgt[lch].sum+sgt[rch].sum)%MOD;
}
void down(int now)
{
        if (sgt[now].l==sgt[now].r)return;
        if (sgt[now].lzy)
        {
                sgt[lch].lzy+=sgt[now].lzy;
                sgt[rch].lzy+=sgt[now].lzy;
                sgt[lch].sum=(sgt[lch].sum+sgt[now].lzy*(sgt[lch].r-sgt[lch].l+1))%MOD;
                sgt[rch].sum=(sgt[rch].sum+sgt[now].lzy*(sgt[rch].r-sgt[rch].l+1))%MOD;
                sgt[now].lzy=0;
        }
}
void Build_sgt(int now,int l,int r)
{
        sgt[now].l=l;
        sgt[now].r=r;
        if (l==r)
        {
                return ;
        }
        Build_sgt(lch,l,(l+r)>>1);
        Build_sgt(rch,((l+r)>>1)+1,r);
}
void Add_sgt(int now,int l,int r)
{
        if (sgt[now].l==l && sgt[now].r==r)
        {
                sgt[now].sum+=sgt[now].r-sgt[now].l+1;
                sgt[now].lzy++;
                return ;
        }
        down(now);
        int mid=(sgt[now].l+sgt[now].r)>>1;
        if (r<=mid)
        {
                Add_sgt(lch,l,r);
        }else if (mid<l)
        {
                Add_sgt(rch,l,r);
        }else
        {
                Add_sgt(lch,l,mid);
                Add_sgt(rch,mid+1,r);
        }
        up(now);
}
int Qry_sgt(int now,int l,int r)
{
        if (sgt[now].l==l && sgt[now].r==r)
        {
                return sgt[now].sum%MOD;
        }
        int mid=(sgt[now].l+sgt[now].r)>>1;
        down(now);
        if (r<=mid)
                return Qry_sgt(lch,l,r);
        if (mid<l)
                return Qry_sgt(rch,l,r);
        return (Qry_sgt(lch,l,mid)+Qry_sgt(rch,mid+1,r))%MOD;
}
struct Edge
{
        int np;
        Edge *next;
}E[MAXE],*V[MAXV];
int tope=-1;
void addedge(int x,int y)
{
        E[++tope].np=y;
        E[tope].next=V[x];
        V[x]=&E[tope];
}
int siz[MAXN];
int depth[MAXN];
int son[MAXN];
int fa[MAXN];
int top[MAXN];
int pos[MAXN],dfstime=0;
void dfs1(int now)
{
        Edge *ne;
        int mxsiz=0;
        siz[now]=1;
        for (ne=V[now];ne;ne=ne->next)
        {
                dfs1(ne->np);
                siz[now]+=siz[ne->np];
                if (siz[ne->np]>mxsiz)
                {
                        mxsiz=siz[ne->np];
                        son[now]=ne->np;
                }
        }
}
void dfs2(int now)
{
        Edge *ne;
        pos[now]=++dfstime;
        if (son[now])
        {
                top[son[now]]=top[now];
                dfs2(son[now]);
        }
        for (ne=V[now];ne;ne=ne->next)
        {
                if (ne->np==son[now])continue;
                top[ne->np]=ne->np;
                dfs2(ne->np);
        }
}
int q[MAXN];
struct qur_t
{
        int x,y,z;
        int ans;
}qur[MAXQ];
vector<pair<int,int> > vec[MAXN];
int main()
{
        freopen("input.txt","r",stdin);
        //freopen("output.txt","w",stdout);
        int x,y,z,i,j,k;
        scanf("%d%d",&n,&m);
        for (i=2;i<=n;i++)
        {
                scanf("%d",&x);x++;
                fa[i]=x;
                addedge(x,i);
        }
        dfs1(1);
        top[1]=1;
        dfs2(1);
        Build_sgt(1,1,n);
        for (i=0;i<m;i++)
        {
                scanf("%d%d%d",&x,&y,&z);
                x++;y++;z++;
                qur[i].x=x;
                qur[i].y=y;
                qur[i].z=z;
                vec[x-1].push_back(make_pair(z,0));
                vec[y].push_back(make_pair(z,0));
        }
        vector<pair<int,int> >::iterator it1;
        for (i=0;i<=n;i++)
        {
                sort(vec[i].begin(),vec[i].end());
                it1=unique(vec[i].begin(),vec[i].end());
                while (vec[i].end()!=it1)vec[i].pop_back();
        }
        for (i=0;i<vec[0].size();i++)
        {
                vec[0][i].second=0;
        }
        int ans;
        for (i=1;i<=n;i++)
        {
                x=i;
                while (x!=0)
                {
                        Add_sgt(1,pos[top[x]],pos[x]);
                        x=fa[top[x]];
                }
                for (j=0;j<vec[i].size();j++)
                {
                        x=vec[i][j].first;
                        ans=0;
                        while (x!=0)
                        {
                                ans=(ans+Qry_sgt(1,pos[top[x]],pos[x]))%MOD;
                                x=fa[top[x]];
                        }
                        vec[i][j].second=ans;
                }
        }
        for (i=0;i<m;i++)
        {
                ans=0;
                it1=lower_bound(vec[qur[i].x-1].begin(),vec[qur[i].x-1].end(),make_pair(qur[i].z,0));
                ans-=it1->second;
                it1=lower_bound(vec[qur[i].y].begin(),vec[qur[i].y].end(),make_pair(qur[i].z,0));
                ans+=it1->second;
                printf("%d\n",ans);
        }
}

bzoj 3626: [LNOI2014]LCA 离线+树链剖分

3626: [LNOI2014]LCA

Description

Input

Output

Sample Input

Sample Output

HINT