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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 201314using 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;}

 

bzoj 3626 LCA