【洛谷 P4211】[LNOI2014]LCA（树链剖分，差分）

题目链接

看到题目肯定首先想到要求LCA(其实是我菜)，可乍一看，n与q的规模为5W，
求LCA的复杂度为\(O(logN)\)，那么总时间复杂度为\(O(nq\ log\ n)\)。
怎么搞呢？
会树上差分的都知道，要对一条链进行操作，比如说链上的节点权值\(+p\)，就要对两个端点分别\(+p\)，然后对\(LCA\)及其父亲分别\(-p\)。
和这个思想差不多，设两个点为\(u, v\)，那么求\(dep(LCA(u,v))\)只需要把\((root,u)\)之前的路径所有点权值\(+1\)，然后统计\((root,v)\)路径上的权值和就行了，具体请自行脑补我觉得很好理解啊，这个过程显然可以用树剖来完成。
然后就是差分，注意不是树上差分，\(\sum_{i=l}^{r}LCA(i,z)=\sum_{i=1}^{r}LCA(i,z)-\sum_{i=1}^{l-1}LCA(i,z)\)
考虑离线，设每个询问为\((l,r,z)\)，定义一个结构体，包含\(x\)(端点)，\(k\)(-1为左端点，1为右端点),\(z\)(要求LCA的点),\(id\)(是哪个询问)，用2个空间分别存储左端点\(l\)和右端点\(r\)，然后按端点编号排序。此时存放询问的数组应该是有\(2q\)个的。定义一个指针\(cur\)，表示当前已经把前\(cur\)个点到根的路径\(+1\)了。枚举\(i(1~2q)\)，如果是左端点，把\(cur\)更新到\(x-1\)，(\(while(cur<x-1) Updata(1,++cur);\))，把\(ans[id]\)减去\(z\)到根的权值和，相当于\(-\sum_{i=1}^{l-1}LCA(i,z)\)。如果是右端点，同理，把\(cur\)更新到\(x\)，然后\(ans[id]\)加上...这题就愉快地切掉了。时间复杂度\(O(n\ log^2\ n)\)

手写树剖一遍过，好激动

#include <cstdio>
#include <algorithm>
using std::swap;
using std::sort;
#define left (now << 1)
#define right (now << 1 | 1)
const int MAXN = 50010;
const int MOD = 201314;
struct Edge{
    int next, to;
}e[MAXN];
int head[MAXN], num, n, q, a, b, c;
int fa[MAXN], maxson[MAXN], size[MAXN], top[MAXN], dfsx[MAXN], ID, pos[MAXN], dep[MAXN], ans[MAXN];
inline void Add(int from, int to){
    e[++num].to = to;
    e[num].next = head[from];
    head[from] = num;
}
int sum[MAXN << 2], lazy[MAXN << 2];
inline void pushdown(int now, int len){
    if(lazy[now]){
      sum[left] += lazy[now] * (len - (len >> 1));
      sum[right] += lazy[now] * (len >> 1);
      lazy[left] += lazy[now];
      lazy[right] += lazy[now];
      lazy[now] = 0;
    }
}
inline void pushup(int now){
    sum[now] = sum[left] + sum[right];
}
/*void Build(int now, int l, int r){
    if(l != r){
      int mid = (l + r) >> 1;
      Build(left, l, mid);
      Build(right, mid + 1, r);
      pushup(now);
    }
    else sum[now] = a[l];
}*/
void Change(int now, int l, int r, int wl, int wr, int p){
    if(l > wr || r < wl) return ;
    if(l >= wl && r <= wr){
      sum[now] += p * (r - l + 1); lazy[now] += p; return ;
    }
    pushdown(now, r - l + 1);
    int mid = (l + r) >> 1;
    Change(left, l, mid, wl, wr, p);
    Change(right, mid + 1, r, wl, wr, p);
    pushup(now);
}
int Query(int now, int l, int r, int wl, int wr){
    if(l > wr || r < wl) return 0;
    if(l >= wl && r <= wr) return sum[now];
    pushdown(now, r - l + 1);
    int ans = 0;
    int mid = (l + r) >> 1;
    ans += Query(left, l, mid, wl, wr);
    ans += Query(right, mid + 1, r, wl, wr);
    return ans;
}
struct ASK{
    int x, z, k, id;
    bool operator < (const ASK &A) const{  //按照端点排序
        return x == A.x ? k < A.k : x < A.x;
    }
}Ans[MAXN << 1];
void dfs1(int u){
    dep[u] = dep[fa[u]] + 1;
    size[u] = 1;
    for(int i = head[u]; i; i = e[i].next){
       dfs1(e[i].to);
       size[u] += size[e[i].to];
       if(size[e[i].to] > size[maxson[u]])
         maxson[u] = e[i].to;
    }
}
void dfs2(int u, int t){
    dfsx[++ID] = u;
    top[u] = t;
    if(maxson[u]) dfs2(maxson[u], t);
    for(int i = head[u]; i; i = e[i].next){
       if(e[i].to != maxson[u])
         dfs2(e[i].to, e[i].to);
    }
}
void Updata(int u, int v, int p){
    int x = top[u], y = top[v];
    while(x != y){
      if(dep[x] > dep[y]) swap(x, y), swap(u, v);
      Change(1, 1, n, pos[y], pos[v], p);
      u = fa[x]; x = top[u];
      v = fa[y]; y = top[v];
    }
    if(dep[u] > dep[v]) swap(u, v);
    Change(1, 1, n, pos[u], pos[v], p);
}
int Solve(int u, int v){
    int ans = 0;
    int x = top[u], y = top[v];
    while(x != y){
      if(dep[x] > dep[y]) swap(x, y), swap(u, v);
      ans += Query(1, 1, n, pos[y], pos[v]);
      u = fa[x]; x = top[u];
      v = fa[y]; y = top[v];
    }
    if(dep[u] > dep[v]) swap(u, v);
    ans += Query(1, 1, n, pos[u], pos[v]);
    return ans;
}
int cur;
int main(){
    scanf("%d%d", &n, &q);
    for(int i = 2; i <= n; ++i)
       scanf("%d", &fa[i]), Add(fa[i] += 1, i);
    for(int i = 1; i <= q; ++i){
       scanf("%d%d%d", &a, &b, &c);
       Ans[i].x = a + 1;
       Ans[i + q].x = b + 1;
       Ans[i].z = Ans[i + q].z = c + 1;
       Ans[i].k = -1;
       Ans[i + q].k = 1;
       Ans[i].id = Ans[i + q].id = i;
    }
    dfs1(1);
    dfs2(1, 1);
    for(int i = 1; i <= n; ++i)
       pos[dfsx[i]] = i;
    sort(Ans + 1, Ans + q * 2 + 1);
    for(int i = 1; i <= 2 * q; ++i){
       if(Ans[i].k < 0)  //是左端点
         while(cur < Ans[i].x - 1)
           Updata(1, ++cur, 1);
       else    //是右端点
         while(cur < Ans[i].x)
           Updata(1, ++cur, 1);
       ans[Ans[i].id] += Ans[i].k * Solve(1, Ans[i].z);
    }
    for(int i = 1; i <= q; ++i)
       printf("%d\n", (ans[i] + MOD) % MOD);
    return 0;
}