题解 LNOI2014 LCA

题目：传送门

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这道题根本不用lca，也没有部分分。。。

考虑求两个点xy的lca的深度。

我们将x到树根所有点的值都加1，然后查询y到根的和，其实就是lca的深度。

所以本题离线一下上树剖乱搞就可以了。

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AC代码如下：
718ms 17348Kib

 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <algorithm>

 using namespace std;

 namespace StandardIO {

     template<typename T> inline void read (T &x) {
         x=;T f=;char c=getchar();
         for (; c<''||c>''; c=getchar()) if (c=='-') f=-;
         for (; c>=''&&c<=''; c=getchar()) x=x*+c-'';
         x*=f;
     }
     template<typename T> inline void write (T x) {
         if (x<) putchar('-'),x=-x;
         if (x>=) write(x/);
         putchar(x%+'');
     }

 }

 using namespace StandardIO;

 namespace Solve {

     const int MOD=;
     const int N=;

     struct Tree {
         int tree[N*],tag[N*];
         void pushdown (int pos,int left,int right) {
             if (tag[pos]) {
                 int mid=(left+right)/;
                 tree[pos*]+=(mid-left+)*tag[pos],tree[pos*]%=MOD;
                 tree[pos*+]+=(right-mid)*tag[pos],tree[pos*+]%=MOD;
                 tag[pos*]+=tag[pos],tag[pos*+]+=tag[pos],tag[pos*]%=MOD,tag[pos*+]%=MOD;
                 tag[pos]=;
             }
         }
         void pushup (int pos) {
             tree[pos]=tree[pos*]+tree[pos*+],tree[pos]%=MOD;
         }
         void update (int pos,int left,int right,int L,int R,int add) {
             if (L<=left&&right<=R) {
                 tree[pos]+=add*(right-left+),tree[pos]%=MOD;
                 tag[pos]+=add,tag[pos]%=MOD;
                 return;
             }
             pushdown(pos,left,right);
             int mid=(left+right)/;
             if (L<=mid) update(pos*,left,mid,L,R,add);
             if (R>mid) update(pos*+,mid+,right,L,R,add);
             pushup(pos);
         }
         int query (int pos,int left,int right,int L,int R) {
             if (L<=left&&right<=R) return tree[pos];
             pushdown(pos,left,right);
             int mid=(left+right)/;
             int ans=;
             if (L<=mid) ans+=query(pos*,left,mid,L,R),ans%=MOD;
             if (R>mid) ans+=query(pos*+,mid+,right,L,R),ans%=MOD;
             return ans;
         }
     } ljz;
     int n,q;
     vector<int>M[N];
     int dep[N],siz[N],fa[N],son[N];
     int ind[N],cnt;
     int top[N];

     void dfs1 (int now,int father) {
         dep[now]=dep[father]+,fa[now]=father,siz[now]=;
         for (register vector<int>::iterator i=M[now].begin(); i!=M[now].end(); i++) {
             if(*i==father) continue;
             dfs1(*i,now);
             siz[now]+=siz[*i];
             if (siz[*i]>siz[son[now]]) son[now]=*i;
         }
     }
     void dfs2(int now,int tp){
         top[now]=tp,ind[now]=++cnt;
         if (son[now]) dfs2(son[now],tp);
         for (register vector<int>::iterator i=M[now].begin(); i!=M[now].end(); i++) {
             if (*i==fa[now]||*i==son[now]) continue;
             dfs2(*i,*i);
         }
     }
     void upd (int x,int add){
         while (x) {
             ljz.update(,,n,ind[top[x]],ind[x],add);
             x=fa[top[x]];
         }
     }
     int que (int x) {
         int ans=;
         while (x) {
             ans+=ljz.query(,,n,ind[top[x]],ind[x]),ans%=MOD;
             x=fa[top[x]];
         }
         return ans;
     }
     int A[N];
     struct Q{
         int val,ind;
         Q () {val=ind=;}
         Q (int a,int b) :val(a),ind(b) {}
         friend bool operator < (Q a,Q b) {
             return a.val<b.val;
         }
     } s1[N],s2[N];
     int a1=,a2=;
     int Ans[N];

     inline void solve () {
         read(n),read(q);
         for (register int i=; i<=n; i++) {
             int a;
             read(a);
             M[a+].push_back(i);
         }
         dfs1(,);
         dfs2(,);
         for (register int i=; i<=q; i++) {
             int a,b,c;
             read(a),read(b),read(c);
             a++,b++,c++;
             A[i]=c;
             s1[i]=Q(a-,i),s2[i]=Q(b,i);
         }
         sort(s1+,s1+q+);
         sort(s2+,s2+q+);
         while (s1[a1].val==) a1++;
         for (register int i=; i<=n; i++) {
             upd(i,);
             while (s1[a1].val==i) {
                 Ans[s1[a1].ind]-=que(A[s1[a1].ind])-MOD,Ans[s1[a1].ind]%=MOD;
                 a1++;
             }
             while (s2[a2].val==i) {
                 Ans[s2[a2].ind]+=que(A[s2[a2].ind]),Ans[s2[a2].ind]%=MOD;
                 a2++;
             }
         }
         for (register int i=; i<=q; i++) {
             write(Ans[i]),putchar('\n');
         }
     }

 }

 using namespace Solve;

 int main () {
 //    freopen(".in","r",stdin);
 //    freopen(".out","w",stdout);
     solve();
 }