【LNOI 2014】 LCA

【题目链接】

         点击打开链接

【算法】

           考虑求lca(x,y)的深度

           我们可以将从根到x路径上的点都打上标记，然后，询问y到根上路径的权值和

           那么，求sigma(depth(lca(i,z)))（l <= i <= r )，我们可以将区间[l,r]中的点依次打上标记，然后，询问点z到根路径

           上的权值和

           因为此题有多组询问，显然在线很难做，因此，我们考虑离线计算答案

           求sigma(depth(lca(i,z))) (l <= i <= r)，我们可以转化为

           sigma(depth(lca(i,z))) ( 0 <= i <= r) - sigma(depth(lca(i,z))) (0 <= i <= l - 1)

           那么，树链剖分/动态树都可以解决这道题，树链剖分的时间复杂度是O((n + q) log(n)^2)的，而动态树的时间复杂度是             O((n + q) log(n))的

【代码】

           由于笔者太弱，不会动态树，所以这份代码是树链剖分的写法

         

#include<bits/stdc++.h>
using namespace std;
#define MAXM 50010
const int P = 201314;

struct Edge
{
    int to,nxt;
} e[MAXM];
struct Query
{
    int pos,opt,z,id;    
} q[MAXM*2];

int i,n,m,f,timer,tot,cnt,now,l,r,z;
int dfn[MAXM],size[MAXM],top[MAXM],head[MAXM],son[MAXM],ans[MAXM],fa[MAXM];

struct SegmentTree
{
        struct Node
        {
                int l,r;
                int sum,tag;
        } Tree[MAXM*4];
        inline void build(int index,int l,int r)
        {
                int mid;
                Tree[index].l = l; Tree[index].r = r;
                Tree[index].sum = Tree[index].tag = 0;
                if (l == r) return;
                mid = (l + r) >> 1;
                build(index<<1,l,mid);
                build(index<<1|1,mid+1,r);
        }
        inline void pushdown(int index)
        {
                int l = Tree[index].l,r = Tree[index].r;
                int mid = (l + r) >> 1;
                if (Tree[index].tag)
                {
                        Tree[index<<1].sum = (Tree[index<<1].sum + (mid - l + 1) * Tree[index].tag) % P;
                        Tree[index<<1|1].sum = (Tree[index<<1|1].sum + (r - mid) * Tree[index].tag) % P;
                        Tree[index<<1].tag = (Tree[index<<1].tag + Tree[index].tag) % P;
                        Tree[index<<1|1].tag = (Tree[index<<1|1].tag + Tree[index].tag) % P;
                        Tree[index].tag = 0;
                }
        }
        inline void update(int index)
        {
                Tree[index].sum = (Tree[index<<1].sum + Tree[index<<1|1].sum) % P;
        }
        inline void modify(int index,int l,int r,int val)
        {
                int mid;
                if (Tree[index].l == l && Tree[index].r == r)
                {
                        Tree[index].sum = (Tree[index].sum + (r - l + 1) * val) % P;
                        Tree[index].tag = (Tree[index].tag + val) % P;
                        return;
                }
                pushdown(index);
                mid = (Tree[index].l + Tree[index].r) >> 1;
                if (mid >= r) modify(index<<1,l,r,val);
                else if (mid + 1 <= l) modify(index<<1|1,l,r,val);
                else
                {
                        modify(index<<1,l,mid,val);
                        modify(index<<1|1,mid+1,r,val);
                }
                update(index);
        }
        inline int query(int index,int l,int r)
        {
                int mid;
                if (Tree[index].l == l && Tree[index].r == r) return Tree[index].sum;
                pushdown(index);
                mid = (Tree[index].l + Tree[index].r) >> 1;
                if (mid >= r) return query(index<<1,l,r);
                else if (mid + 1 <= l) return query(index<<1|1,l,r);
                else return (query(index<<1,l,mid) + query(index<<1|1,mid+1,r)) % P;
        }
} T;
inline bool cmp(Query a,Query b) 
{
    return a.pos < b.pos;
}
inline void add(int u,int v)
{
        tot++;
        e[tot] = (Edge){v,head[u]};
        head[u] = tot;
}
inline void dfs1(int u)
{
        int i,v;
        size[u] = 1;
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                dfs1(v);
                size[u] += size[v];
                if (size[v] > size[son[u]] || !son[u]) son[u] = v;
        }
}
inline void dfs2(int u,int tp)
{
        int i,v;
        top[u] = tp;
        dfn[u] = ++timer;
        if (son[u]) dfs2(son[u],tp);     
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                if (son[u] != v) dfs2(v,v);
        }
}
inline void modify(int pos)
{
        int tp = top[pos];
        while (tp)
        {
                T.modify(1,dfn[tp],dfn[pos],1);
                pos = fa[tp]; tp = top[pos];
        }
        T.modify(1,1,dfn[pos],1);
}
inline int query(int pos)
{
        int tp = top[pos],ans = 0;
        while (tp)
        {
                ans = (ans + T.query(1,dfn[tp],dfn[pos])) % P;
                pos = fa[tp]; tp = top[pos];
        }
        ans = (ans + T.query(1,1,dfn[pos])) % P;
        return ans;
}

int main()
{
    
        scanf("%d%d",&n,&m);
        for (i = 1; i < n; i++) 
        {
                scanf("%d",&fa[i]);
                add(fa[i],i);    
        }
        dfs1(0);
        dfs2(0,0);
        T.build(1,1,timer);
        for (i = 1; i <= m; i++) 
        {
                scanf("%d%d%d",&l,&r,&z);
                if (l != 0) q[++cnt] = (Query){l-1,-1,z,i};
                q[++cnt] = (Query){r,1,z,i};
        }
        sort(q+1,q+cnt+1,cmp);
        now = -1; 
        for (i = 1; i <= cnt; i++)
        {
                while (now + 1 <= q[i].pos) 
                {
                        now++;
                        modify(now);
                } 
                if (q[i].opt == 1) ans[q[i].id] = (ans[q[i].id] + query(q[i].z)) % P;
                else ans[q[i].id] = (ans[q[i].id] - query(q[i].z) + P) % P;
        }
        for (i = 1; i <= m; i++) printf("%d
",ans[i]);
        
        return 0;
}