BZOJ3653 谈笑风生

先对树DFS一边，建立dfs序列

tr[p].dep表示p的深度，tr[p].sz表示p的子树的大小（不包括p自己）

对dfs序列建立主席树乱搞= =

Claris大爷："线段树中区间[a,b]表示深度在[a,b]范围内的sz的和"

貌似明白了>_<

Claris大爷又曰："查询x，y的答案 = tr[x].sz * min(tr[x].dep, y) + dfs序在tr[x].L + 1到tr[x].R之间且深度在tr[x].dep + 1到tr[x].dep + k之间的sz的和"

于是就完了2333

/**************************************************************
    Problem: 3653
    User: rausen
    Language: C++
    Result: Accepted
    Time:9060 ms
    Memory:111928 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
using namespace std;
typedef long long ll;
const int N = 300005;
const int M = 5700000;
 
struct edges {
    int next, to;
    edges() {}
    edges(int _next, int _to) : next(_next), to(_to) {}
}e[N << 1];
 
struct segment_tree {
    ll val;
    int lson, rson;
}seg[M];
 
struct tree_node {
    int fa, dep, sz, L, R;
}tr[N];
 
int n, Q;
int D;
int first[N], tot, cnt;
int q[N], TOT, root[N];
 
inline int read() {
    int x = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
inline void Add_Edges(int x, int y) {
    e[++tot] = edges(first[x], y), first[x] = tot;
    e[++tot] = edges(first[y], x), first[y] = tot;
}
 
void dfs(int p) {
    int x, y;
    tr[p].L = ++cnt;
    q[cnt] = p;
    for (x = first[p]; x; x = e[x].next)
        if ((y = e[x].to) != tr[p].fa) {
            tr[y].fa = p;
            tr[y].dep = tr[p].dep + 1;
            dfs(y);
            tr[p].sz += tr[y].sz + 1;
        }
    tr[p].R = cnt;
}
 
int build(int p, int l, int r, int d, int sz) {
    int y = ++TOT;
    seg[y].val = seg[p].val + sz;
    if (l == r) return y;
    int mid = l + r >> 1;
    if (d <= mid)
        seg[y].lson = build(seg[p].lson, l, mid, d, sz), seg[y].rson = seg[p].rson;
    else
        seg[y].rson = build(seg[p].rson, mid + 1, r, d, sz), seg[y].lson = seg[p].lson;
    return y;       
}
 
ll query(int p, int l, int r, int L, int R) {
    if (L <= l && r <= R)
        return seg[p].val;
    int mid = l + r >> 1;
    ll res = 0;
    if (L <= mid && seg[p].lson)
        res += query(seg[p].lson, l, mid, L, R);
    if (mid < R && seg[p].rson)
        res += query(seg[p].rson, mid + 1, r, L, R);
    return res;
}
 
inline void print(int x, int y) {
    ll a = (ll) tr[x].sz * min(tr[x].dep, y),
       b = query(root[tr[x].R], 0, D, tr[x].dep, tr[x].dep + y),
       c = query(root[tr[x].L], 0, D, tr[x].dep, tr[x].dep + y);
    printf("%lld
", (ll) a + b - c);
}
 
int main() {
    int i, X, Y;
    n = read(), Q = read();
    for (i = 1; i < n; ++i) {
        X = read(), Y = read();
        Add_Edges(X, Y);
    }
    dfs(1);
    for (i = 1; i <= n; ++i)
        D = max(D, tr[i].dep);
    for (i = 1; i <= n; ++i)
        root[i] = build(root[i - 1], 0, D, tr[q[i]].dep, tr[q[i]].sz);
    while (Q--) {
        X = read(), Y = read();
        print(X, Y);
    }
    return 0;
}

（p.s 竟然1WA，蒟蒻简直了。。。还查不错来，要了数据才发现我是沙茶，运算的时候要记得加上"(ll)"）