题目

P3258 [JLOI2014]松鼠的新家

解析

非常裸的一道树剖题

链上修改+单点查询的板子

记录一下所经过的点\(now[i]\)，每次更新\(now[i-1]到now[i]\)

我们链上更新时上一次到的终点，是这一次一次更新的起点，又因为在\(a_n\)处可以不放糖，所以我们每次链上更新完成后，在上条链的终点位置处糖数\(-1\)。

然后套板子直接做

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
int n, m, num, cnt;
int head[N], size[N], f[N], top[N], son[N], dep[N], now[N], id[N];
class node {
    public :
        int v, nx;
} e[N];
class tree {
    public :   
        int sum, lazy;
        int len;
} t[N];

#define lson rt << 1
#define rson rt << 1 | 1

template<class T>inline void read(T &x) {
    x = 0; int f = 0; char ch = getchar();
    while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    x = f ? -x : x;
    return ;
}

inline void add(int u, int v) {
    e[++num].nx = head[u], e[num].v = v, head[u] = num;
}

void dfs1(int u, int fa) {
    size[u] = 1;
    for (int i = head[u]; ~i; i = e[i].nx) {
        int v = e[i].v;
        if (v != fa) {
            dep[v] = dep[u] + 1;
            f[v] = u;
            dfs1(v, u);
            size[u] += size[v];
            if (size[v] > size[son[u]]) son[u] = v;
        }
    }
}

void dfs2(int u, int t) {
    id[u] = ++cnt;
    
    top[u] = t;
    if (son[u]) dfs2(son[u], t);
    for (int i = head[u]; ~i; i = e[i].nx) {
        int v = e[i].v;
        if (v != f[u] && v != son[u]) dfs2(v, v);           //WRONG
    }
}

inline void pushup(int rt) {
    t[rt].sum = t[lson].sum + t[rson].sum;
}

void build(int l, int r, int rt) {
    t[rt].len = r - l + 1;
    if (l == r) return;
    int m = (l + r) >> 1;
    build(l, m, lson);
    build(m + 1, r, rson);
}

inline void pushdown(int rt) {
    if (t[rt].lazy) {
        t[lson].lazy += t[rt].lazy;
        t[rson].lazy += t[rt].lazy;
        t[lson].sum += t[rt].lazy * t[lson].len;
        t[rson].sum += t[rt].lazy * t[rson].len;
        t[rt].lazy = 0;
    }
}

void update(int L, int R, int c, int l, int r, int rt) {
    if (L <= l && r <= R) {
        t[rt].sum += (t[rt].len * c);
        t[rt].lazy += c;
        return;
    }
    pushdown(rt);
    int m = (l + r) >> 1;
    if (L <= m) update(L, R, c, l, m, lson);
    if (R > m) update(L, R, c, m + 1, r, rson);
    pushup(rt);
}

int query(int L, int R, int l, int r, int rt) {
    if (L <= l && r <= R) return t[rt].sum;
    pushdown(rt);
    int m = (l + r) >> 1, ans = 0;
    if (L <= m) ans += query(L, R, l, m, lson);
    if (R > m) ans += query(L, R, m + 1, r, rson);
    return ans;
}

void update_chain(int x, int y, int z) {
    int fx = top[x], fy = top[y];
    while (fx != fy) {
        if (dep[fx] < dep[fy]) swap(fx, fy), swap(x, y);
        update(id[fx], id[x], z, 1, cnt, 1);
        x = f[fx], fx = top[x];
    }
    if (id[x] > id[y]) swap(x, y);
    update(id[x], id[y], z, 1, cnt, 1);
}

int main() {
    memset(head, -1, sizeof(head));
    read(n);
    for (int i = 1, x; i <= n; ++i) read(x), now[i] = x;
    for (int i = 1, x, y; i < n; ++i) read(x), read(y), add(x, y), add(y, x);
    f[1] = 0, dep[1] = 0;
    dfs1(1, 0);
    dfs2(1, 1);
    build(1, n, 1);
    for (int i = 2; i <= n; ++i) update_chain(now[i - 1], now[i], 1), update_chain(now[i], now[i], -1);
    for (int i = 1; i <= n; ++i) printf("%d\n", query(id[i], id[i], 1, n, 1));
    return 0;
}