bzoj3589 动态树 树链剖分+容斥

标签：problem   https   long   stdin   zoj   sum   相加   直接   swa   

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=3589

题解

事件 \(0\) 不需要说，直接做就可以了。

事件 \(1\) 的话，考虑如果直接查询然后相加的话，会有很多段被算重了。于是考虑容斥，把算重的段给减掉就可以了。至于如何计算每一段的答案，直接树剖吧。

时间复杂度 \(O(^5 q\log^2n)\)。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

#define lc o << 1
#define rc o << 1 | 1
#define lowbit(x) ((x) & -(x))

const int N = 200000 + 7;

int n, m, k, dfc;
int dep[N], f[N], siz[N], son[N], dfn[N], pre[N], top[N];
int qx[40], qy[40], pcnt[40];

struct Edge { int to, ne; } g[N << 1]; int head[N], tot;
inline void addedge(int x, int y) { g[++tot].to = y, g[tot].ne = head[x], head[x] = tot; }
inline void adde(int x, int y) { addedge(x, y), addedge(y, x); }

struct Node { int add, sum; } t[N << 2];
inline void qadd(int o, int L, int R, int l, int r, int k) {
    if (l <= L && R <= r) return t[o].add += k, t[o].sum += (R - L + 1) * k, (void)0;
    int M = (L + R) >> 1;
    if (l <= M) qadd(lc, L, M, l, r, k);
    if (r > M) qadd(rc, M + 1, R, l, r, k);
    t[o].sum = t[lc].sum + t[rc].sum + t[o].add * (R - L + 1);
}
inline int qsum(int o, int L, int R, int l, int r, int add = 0) {
    if (l <= L && R <= r) return t[o].sum + add * (R - L + 1);
    int M = (L + R) >> 1;
    if (r <= M) return qsum(lc, L, M, l, r, add + t[o].add);
    if (l > M) return qsum(rc, M + 1, R, l, r, add + t[o].add);
    return qsum(lc, L, M, l, r, add + t[o].add) + qsum(rc, M + 1, R, l, r, add + t[o].add);
}

inline void upd(int x, int k) { qadd(1, 1, n, dfn[x], dfn[x] + siz[x] - 1, k); }
inline int qry(int x, int y) {
    int ans = 0;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) std::swap(x, y);
        ans += qsum(1, 1, n, dfn[top[x]], dfn[x]);
        x = f[top[x]];
    }
    if (dep[x] > dep[y]) std::swap(x, y);
    return ans += qsum(1, 1, n, dfn[x], dfn[y]);
}

inline void dfs1(int x, int fa = 0) {
    dep[x] = dep[fa] + 1, f[x] = fa, siz[x] = 1;
    for fec(i, x, y) if (y != fa) dfs1(y, x), siz[x] += siz[y], siz[y] > siz[son[x]] && (son[x] = y);
}
inline void dfs2(int x, int pa) {
    top[x] = pa, dfn[x] = ++dfc, pre[dfc] = x;
    if (!son[x]) return; dfs2(son[x], pa);
    for fec(i, x, y) if (y != f[x] && y != son[x]) dfs2(y, y);
}
inline int lca(int x, int y) {
    while (top[x] != top[y]) dep[top[x]] >= dep[top[y]] ? x = f[top[x]] : y = f[top[y]];
    return dep[x] < dep[y] ? x : y;
}

inline bool intree(int x, int y) { return dfn[y] >= dfn[x] && dfn[y] <= dfn[x] + siz[x] - 1; }
inline pii merge(pii l1, pii l2) {
    if (dep[l1.fi] > dep[l1.se]) std::swap(l1.fi, l1.se);
    if (dep[l2.fi] > dep[l2.se]) std::swap(l2.fi, l2.se);
    if (dep[l1.fi] > dep[l2.fi]) std::swap(l1, l2);
    if (intree(l1.fi, l2.fi) && intree(l2.fi, l1.se)) return pii(l2.fi, lca(l1.se, l2.se));
    else return pii(0, 0);
}

inline void work() {
    dfs1(1), dfs2(1, 1);
    read(m);
    while (m--) {
        int opt, x, y;
        read(opt);
        if (opt == 0) read(x), read(y), upd(x, y);
        else {
            read(k);
            for (int i = 1; i <= k; ++i) read(qx[1 << (i - 1)]), read(qy[1 << (i - 1)]);
            int ans = 0, S = (1 << k) - 1;
            for (int s = 1; s <= S; ++s) {
                int sta = s ^ lowbit(s);
                pcnt[s] = pcnt[sta] + 1;
                if (sta) {
                    pii hkk = merge(pii(qx[lowbit(s)], qy[lowbit(s)]), pii(qx[sta], qy[sta]));
                    qx[s] = hkk.fi, qy[s] = hkk.se;
                }
                if (qx[s]) {
                    if (pcnt[s] & 1) ans += qry(qx[s], qy[s]);
                    else ans -= qry(qx[s], qy[s]);
                }
            }
            printf("%d\n", ans & ((1 << 31) - 1));
        }
    }
}

inline void init() {
    read(n);
    int x, y;
    for (int i = 1; i < n; ++i) read(x), read(y), adde(x, y);
}

int main() {
#ifdef hzhkk
    freopen("hkk.in", "r", stdin);
#endif
    init();
    work();
    fclose(stdin), fclose(stdout);
    return 0;
}

bzoj3589 动态树 树链剖分+容斥

标签：problem   https   long   stdin   zoj   sum   相加   直接   swa   

原文地址：https://www.cnblogs.com/hankeke/p/bzoj2589.html