bzoj 3626

http://www.lydsy.com/JudgeOnline/problem.php?id=3626

让我比较惊讶的一道链剖裸题(' '    ) 做法很精妙

首先我们考虑对于单个询问时可以拆分成(1, l - 1, z) 和 (1, r, z) 的， 然后考虑对于每一次询问可以表示为将(1, l) 的所有点到根的全部加1 然后求z到根路径的的和。 所以将询问离线， 按询问的l值排序，每一次遇到新的l值就将这一段的点到根的路径全部加1，然后查询即可

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long ll;
const ll maxn = 100001;
const ll mod = 201314;

ll n, m;

ll int_get() {
    ll x = 0; char c = (char)getchar(); bool f = 0;
    while(!isdigit(c)) {
        if(c == '-') f = 1;
        c = (char)getchar();
    }
    while(isdigit(c)) {
        x = x * 10 + (int)(c - '0');
        c = (char)getchar();
    }
    if(f) x = -x;
    return x;
}

struct seg {
    ll data, lazy; 
    seg *l, *r;
}tr[maxn * 3];ll sege = 0;
seg* root;

void test(seg* x, ll l, ll r) {
    cout << l <<" "<< r <<" "<< x-> data <<" "<< x-> lazy << endl;
    if(l ^ r) {
        ll mid = (l + r) >> 1;
        test(x-> l, l, mid), test(x-> r, mid + 1, r);
    }
}

seg* build(ll l, ll r) {
    seg* x = tr + sege ++;
    if(l ^ r) {
        ll mid = (l + r) >> 1;
        x-> l = build(l, mid);
        x-> r = build(mid + 1, r);
    }
    return x;
}

void update(seg* x) {
    if(x-> l) x-> data = x-> l-> data + x-> r-> data;
}

void pushdown(seg* x, ll l, ll r) {
    if(x-> l && x-> lazy != 0) {
        ll mid = (l + r) >> 1;
        x-> l-> lazy += x-> lazy, x-> l-> data += (mid - l + 1) * x-> lazy;
        x-> r-> lazy += x-> lazy, x-> r-> data += (r - mid) * x-> lazy;
        x-> lazy = 0;
    }
}

void addlazy(seg* x, ll l, ll r, ll ls, ll rs, ll v) {
    if(l == ls && r == rs) x-> data += v * (rs - ls + 1), x-> lazy += v; 
    else {
        pushdown(x, l, r);
        ll mid = (l + r) >> 1;
        if(rs <= mid) addlazy(x-> l, l, mid, ls, rs, v);
        else if(ls > mid) addlazy(x-> r, mid + 1, r, ls, rs, v);
        else addlazy(x-> l, l, mid, ls, mid, v), addlazy(x-> r, mid + 1, r, mid + 1, rs, v);
        update(x);
    }
}

ll ask(seg* x, ll l, ll r, ll ls, ll rs) {
    if(l == ls && r == rs) return x-> data;
    else {
        pushdown(x, l, r);
        ll mid = (l + r) >> 1;
        if(rs <= mid) return ask(x-> l, l, mid, ls, rs); 
        else if(ls > mid) return ask(x-> r, mid + 1, r, ls, rs);
        else return ask(x-> l, l, mid, ls, mid) + ask(x-> r, mid + 1, r, mid + 1, rs);
    }
}

struct edge {
    ll t;
    edge* next;
}e[maxn * 2], *head[maxn]; ll ne = 0;

void addedge(ll f, ll t) {
    e[ne].t = t, e[ne].next = head[f], head[f] = e + ne ++;
}

ll h[maxn], size[maxn], fa[maxn], un[maxn], map[maxn], num = 0;

void dfs(ll x, ll pre) {
    fa[x] = pre, size[x] = 1; h[x] = h[pre] + 1;
    for(edge* p = head[x]; p; p = p-> next) {
        if(p-> t != pre) dfs(p-> t, x), size[x] += size[p-> t]; 
    }
}

void divide(ll x, ll Un) {
    un[x] = Un, map[x] = ++ num;
    if(size[x] == 1) return ;
    ll Max = 0, pos; 
    for(edge* p = head[x]; p; p = p-> next) {
        if(p-> t != fa[x] && size[p-> t] > Max) Max = size[p-> t], pos = p-> t; 
    }
    divide(pos, Un);
    for(edge* p = head[x]; p; p = p-> next) {
        if(p-> t != fa[x] && p-> t != pos) divide(p-> t, p-> t);
    }
}

void add(ll a, ll b, ll v) {
    ll ls, rs;  
    while(un[a] != un[b]) {
        if(h[un[a]] > h[un[b]]) {
            ls = map[un[a]], rs = map[a];
            addlazy(root, 1, n, ls, rs, v); 
            a = fa[un[a]];
        }
        else {
            ls = map[un[b]], rs = map[b];
            addlazy(root, 1, n, ls, rs, v); 
            b = fa[un[b]];
        } 
    }
    ls = map[a], rs = map[b];
    if(ls > rs) swap(ls, rs);
    addlazy(root, 1, n, ls, rs, v);
}

ll get(ll a, ll b) {
    ll ret = 0;
    ll ls, rs;
    while(un[a] != un[b]) {
        if(h[un[a]] > h[un[b]]) {
            ls = map[un[a]], rs = map[a];
            ret += ask(root, 1, n, ls, rs);
            a = fa[un[a]];
        }
        else {
            ls = map[un[b]], rs = map[b];
            ret += ask(root, 1, n, ls, rs); 
            b = fa[un[b]];
        }
    }
    ls = map[a], rs = map[b];
    if(ls > rs) swap(ls, rs);
    ret += ask(root, 1, n, ls, rs);
    return ret;
}

struct block {
    ll x, p, pos, fl;
}o[maxn]; ll oe = 0;

bool cmp(block a, block b) {
    return a. x < b. x;
}

ll ans[maxn];

void read() {
    n = int_get(); m = int_get(); 
    for(ll i = 2; i <= n; ++ i) {
        ll u = int_get() + 1;
        addedge(u, i), addedge(i, u);
    }
    for(ll i = 1; i <= m; ++ i) {
        ll a, b, c;
        a = int_get() + 1, b = int_get() + 1, c = int_get() + 1;
        ++ oe, o[oe]. x = a - 1, o[oe]. p = c, o[oe]. pos = i, o[oe]. fl = -1;
        ++ oe, o[oe]. x = b, o[oe]. p = c, o[oe].pos = i, o[oe].fl = 1;
    }
    dfs(1, 0), divide(1, 1);
    root = build(1, n);
}

void sov() {
    sort(o + 1, o + 1 + oe, cmp); ll pl = 0;
    for(ll i = 1; i <= oe; ++ i) {
        while(pl < o[i]. x) {
            ++ pl; add(1, pl, 1); 
        }
        //test(root, 1, n); cout << endl;
        ans[o[i]. pos] += o[i]. fl * get(1, o[i]. p); 
    }
    for(ll i = 1; i <= m; ++ i) printf("%lld
", (ans[i] % mod + mod)% mod);
}

int main() {
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    read(); 
    sov();
    return 0;
}