BZOJ3626LCA(树剖＋线段树＋LCA+差分)

题面

给定一棵树，有ｑ次询问，每次询问都是l,r,p，每次询问∑i=lrdeep[LCA(i,p)].

数据范围达到了上万，显然每次询问要么log地询问，要么O(1)询问，log级的显然是比较难处理的，于是我们想到把询问离线，发现这东西满足差分的性质，即处理出[1,l−1]和[1,r]然后两者相减就是答案．想到这个之后我们考虑怎么求对于每个询问的答案．我们这么想，我们从１用一个指针往后扫，发现遇见这个点是询问中的点（l-1 or r）然后我们对这个询问的ｐ处理下，算出答案．关键是这个答案要怎么算？这里要用到一个非常巧妙的东西，我们从１往后扫的同时，把这个点到根节点上所有点都打上＋１的标记，然后我们有一个ｐ，实际上答案就是求ｐ到根路径上标记之和，在纸上画一下就好了．这样每次(log2n)2的复杂度，总复杂度为O(n(log2n)2).

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>using namespace std;typedef long long LL;#define REP(i,a,b) for(register int i = (a),i##_end_ = (b); i <= i##_end_; ++i)#define DREP(i,a,b) for(register int i = (a),i##_end_ = (b);i >= i##_end_; --i)#define mem(a,b) memset(a,b,sizeof(a));int read(){    register int flag = 1,sum = 0;char c = getchar();    while(!isdigit(c)) { if(c == '-')flag = -1; c = getchar(); }    while(isdigit(c)) { sum = sum * 10 + c - '0'; c = getchar(); }    return sum * flag;}const int maxn = 400000+10;const int mod = 201314;int n,m;int be[maxn],ne[maxn],to[maxn],e;void add(int x,int y){    to[++e] = y;ne[e] = be[x];be[x] = e;}int dep[maxn],size[maxn],son[maxn],fa[maxn];void dfs1(int x,int dp,int faa){    dep[x] = dp;size[x] = 1;    for(int i = be[x]; i; i = ne[i])    {        int v = to[i];        if(v != faa)        {            dfs1(v,dp+1,x);fa[v] = x;            size[x] += size[v];            if(!son[x] || size[son[x]] < size[v])son[x] = v;        }    }}int top[maxn],tid[maxn],cnt,vis[maxn];void dfs2(int x,int tp){    vis[x] = 1;top[x] = tp;tid[x] = ++cnt;    if(!son[x])return ;    dfs2(son[x],tp);    for(int i = be[x]; i; i = ne[i])    {        int v = to[i];        if(!vis[v])dfs2(v,v);    }}struct T{    int id,flag,z,p;    LL ans;    bool operator < (const T&u)const    {        return z < u.z;    }}q[maxn];int num;struct node{    int l,r,ld,rd,p;}ask[maxn];LL tr[maxn<<2],tag[maxn<<2];void pushdown(int h,int l,int r){    int mid = (l + r) >> 1;    tag[h<<1] += tag[h];    tag[h<<1|1] += tag[h];    (tr[h<<1] += (mid - l + 1) * tag[h])%=mod;    (tr[h<<1|1] += (r - mid) * tag[h])%=mod;    tag[h] = 0;}void updata(int h,int l,int r,int q,int w){    if(q <= l &&r <= w)    {        (tr[h] += w - q + 1)%=mod;        tag[h]++;return ;    }    int mid = (l + r) >> 1;    if(tag[h])pushdown(h,l,r);    if(w <= mid)updata(h<<1,l,mid,q,w);    else if(q > mid)updata(h<<1|1,mid+1,r,q,w);    else updata(h<<1,l,mid,q,w),updata(h<<1|1,mid+1,r,q,w);    tr[h] = tr[h<<1] + tr[h<<1|1];}LL query(int h,int l,int r,int q,int w){    if(q <= l && r <= w)return tr[h];    if(tag[h])pushdown(h,l,r);    int mid = (l + r) >> 1;    if(w <= mid)return query(h<<1,l,mid,q,w);    else if(q > mid)return query(h<<1|1,mid+1,r,q,w);    else return query(h<<1,l,mid,q,w)+query(h<<1|1,mid+1,r,q,w);}void add(int x){    int y = 0;     while(top[x] != top[y])    {        if(dep[top[x]] < dep[top[y]])swap(x,y);        updata(1,1,cnt,tid[top[x]],tid[x]);        x = fa[top[x]];    }    updata(1,1,cnt,tid[0],tid[x]);}void solve(int id){    int x = q[id].p,y = 0;    LL res = 0;    while(top[x]!=top[y])    {        if(dep[top[x]] < dep[top[y]])swap(x,y);        (res += query(1,1,cnt,tid[top[x]],tid[x]))%=mod;        x = fa[top[x]];    }    (res += query(1,1,cnt,tid[0],tid[x]))%=mod;    q[id].ans = res;}int main(){    freopen("1.in","r",stdin);    freopen("1.out","w",stdout);    n = read();m = read();    REP(i,1,n-1)    {        int x = read();        add(x,i);add(i,x);    }//slpf    dfs1(0,0,-1);dfs2(0,0);    //sort,prehandle query    REP(i,1,m)    {        ask[i].l = read(),ask[i].r = read(),ask[i].p = read();        q[++num] = (T){i,0,ask[i].l-1,ask[i].p};        q[++num] = (T){i,1,ask[i].r,ask[i].p};    }    sort(q+1,q+1+num);      REP(i,1,num)    {        if(!q[i].flag)ask[q[i].id].ld = i;        else ask[q[i].id].rd = i;    }    int ths = 1,pt = -1;    for(;pt < n && ths <= num;++ths)    {        while(pt < q[ths].z)            ++pt,add(pt);        solve(ths);    }    REP(i,1,m)        cout<<(q[ask[i].rd].ans-q[ask[i].ld].ans+mod)%mod<<endl;}