bzoj 3626 LCA

　　　　这一道题咋一看只觉得是离线，可以求出所有的ｆ（１，ｉ，ｚ），　答案就等于ｆ（１，ｒ，ｚ）－ｆ（１，ｌ－１，ｚ）。但是没有具体的做法，但是求ＬＣＡ的深度和有一个非常巧妙的做法，每加一个点，就把这个点到根的路径上的点权值＋１，这样计算某个点和之前所有点ＬＣＡ深度和就可以统计这个点到根的路径上的点的权值和。这样就可以用树链剖分很快的修改和得出答案，这题就解决了。

　　　　上代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <queue>
#define N 51000
#define yu 201314
using namespace std;

struct sss
{
    int place, askp;
    int num, nump;
}ask[N*2];
struct ss
{
    int num, push;
}t[N*4];
int n, m, nowplace = 0;
int p[N], v[N], next[N], bnum = 0;
int ans[N][2] = {0};
int fa[N], deep[N], siz[N], son[N], w[N], top[N];

bool cmp(sss x, sss y) { return x.place < y.place; }

void addbian(int x, int y)
{
    bnum++; next[bnum] = p[x]; p[x] = bnum; v[bnum] = y;
}

void build_tree(int now, int l, int r)
{
    t[now].num = 0; t[now].push = 0;
    if (l == r) return; int mid = (l+r)/2;
    build_tree(now*2, l, mid); build_tree(now*2+1, mid+1, r);
}

void dfs_1(int now, int fat, int de)
{
    int k = p[now]; fa[now] = fat; deep[now] = de;
    int maxsonnum = 0; siz[now] = 1; son[now] = 0;
    while (k)
    {
        if (v[k] != fat)
        {
            dfs_1(v[k], now, de+1);
            siz[now] += siz[v[k]];
            if (siz[v[k]] > maxsonnum)
            {
                maxsonnum = siz[v[k]];
                son[now] = v[k];
            }
        }
        k = next[k];
    }
    return;
}

void dfs_2(int now, int fat, int nowtop)
{
    int k = p[now]; top[now] = nowtop; w[now] = ++nowplace;
    if (son[now]) dfs_2(son[now], now, nowtop);
    while (k)
    {
        if (v[k] != son[now] && v[k] != fat)
            dfs_2(v[k], now, v[k]);
        k = next[k];
    }
    return;
}

void downdate(int now, int l, int r)
{
    if (!t[now].push) return; int mid = (l+r)/2;
    t[now*2].push += t[now].push;
    t[now*2+1].push += t[now].push;
    t[now*2].num += (mid-l+1) * t[now].push;
    t[now*2+1].num += (r-mid) * t[now].push;
    if (t[now*2].num > yu) t[now*2].num %= yu;
    if (t[now*2+1].num > yu) t[now*2+1].num %= yu;
    t[now].push = 0; return;
}

void tadd(int now, int l, int r, int al, int ar)
{
    if (al <= l && r <= ar)
    {
        t[now].num += r-l+1;
        if (t[now].num > yu) t[now].num %= yu;
        t[now].push ++; return;
    }
    int mid = (l+r)/2; downdate(now, l, r);
    if (al <= mid) tadd(now*2, l, mid, al, ar);
    if (ar > mid) tadd(now*2+1, mid+1, r, al, ar);
    t[now].num = t[now*2].num + t[now*2+1].num;
    if (t[now].num > yu) t[now].num %= yu;
}

int task(int now, int l, int r, int al, int ar)
{
    if (al <= l && r <= ar) return t[now].num;
    int mid = (l+r)/2, zans = 0; downdate(now, l, r);
    if (al <= mid) zans = task(now*2, l, mid, al, ar);
    if (ar > mid) zans += task(now*2+1, mid+1, r, al, ar);
    if (zans > yu) zans %= yu;
    return zans;
}

int askk(int u, int v)
{
    int f1 = top[u], f2 = top[v];
    if (deep[f1] < deep[f2]) { swap(f1, f2); swap(u, v); }
    if (f1 == f2)
    {
        if (u == v) return task(1, 1, n, w[u], w[u]);
        return task(1, 1, n, min(w[u], w[v]), max(w[u], w[v]));
    }
    int zans = task(1, 1, n, w[f1], w[u]);
    zans += askk(fa[f1], v); if (zans > yu) zans %= yu;
    return zans;
}

void add(int u, int v)
{
    int f1 = top[u], f2 = top[v];
    if (deep[f1] < deep[f2]) { swap(f1, f2); swap(u, v); }
    if (f1 == f2)
    {
        if (u == v) tadd(1, 1, n, w[u], w[u]);
        else tadd(1, 1, n, min(w[u], w[v]), max(w[u], w[v]));
        return;
    }
    tadd(1, 1, n, w[f1], w[u]); add(fa[f1], v);
}

int main()
{
    scanf("%d%d", &n, &m); build_tree(1, 1, n);
    for (int i = 1; i < n; ++i)
    {
        int x; scanf("%d", &x);
        addbian(x+1, i+1);
    }
    dfs_1(1, 0, 1); dfs_2(1, 0, 1);
    for (int i = 1; i <= m; ++i)
    {
        int x, y, z; scanf("%d%d%d", &x, &y, &z); x++; y++; z++;
        ask[i*2-1].place = x-1; ask[i*2-1].askp = z;
        ask[i*2-1].num = i; ask[i*2-1].nump = 0;
        ask[i*2].place = y; ask[i*2].askp = z;
        ask[i*2].num = i; ask[i*2].nump = 1;
    }
    sort(ask+1, ask+1+2*m, cmp); int nowplace = 0;
    for (int i = 1; i <= m*2; ++i)
    {
        while (ask[i].place > nowplace)
        {
            nowplace++;
            add(1, nowplace);
        }
        if (ask[i].place)
            ans[ask[i].num][ask[i].nump] = askk(1, ask[i].askp);
        else ans[ask[i].num][ask[i].nump] = 0;
    }
    for (int i = 1; i <= m; ++i)
        printf("%d\n", (ans[i][1]+yu-ans[i][0]) % yu);
    return 0;
}