【题解】51nod-2004-终结之时

题目

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=2004

题意

在支配树上实现一些持久化操作,详见题面。

题解

  • 支配树裸题 + 树链剖分裸题 + 持久化线段树裸题
  • 求若干个点到根的链上结点的并的原理和求虚树很相似,只需要考虑 dfs 序相邻的结点。

代码(需要一些空间优化技巧)

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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int N = 5E4 + 100;

vector<int> G[N];

namespace tl{
vector<int> G[N], rG[N];
int fa[N], idx[N], clk, ridx[N];
int c[N], best[N], semi[N], idom[N];
void init(int n) {
clk = 0;
fill(c, c + n + 1, -1);
FOR (i, 1, n + 1) ::G[i].clear();
FOR (i, 1, n + 1) semi[i] = best[i] = i;
fill(idx, idx + n + 1, 0);
}
void dfs(int u) {
idx[u] = ++clk; ridx[clk] = u;
for (int& v: G[u]) if (!idx[v]) { fa[v] = u; dfs(v); }
}
int fix(int x) {
if (c[x] == -1) return x;
int &f = c[x], rt = fix(f);
if (idx[semi[best[x]]] > idx[semi[best[f]]]) best[x] = best[f];
return f = rt;
}
void go(int rt) {
dfs(rt);
FORD (i, clk, 1) {
int x = ridx[i], mn = clk + 1;
for (int& u: rG[x]) {
if (!idx[u]) continue;
fix(u); mn = min(mn, idx[semi[best[u]]]);
}
c[x] = fa[x];
::G[semi[x] = ridx[mn]].push_back(x);
x = ridx[i - 1];
for (int& u: ::G[x]) {
fix(u);
if (semi[best[u]] != x) idom[u] = best[u];
else idom[u] = x;
}
::G[x].clear();
}

FOR (i, 2, clk + 1) {
int u = ridx[i];
if (idom[u] != semi[u]) idom[u] = idom[idom[u]];
::G[idom[u]].push_back(u);
}
}
}

int fa[N], dep[N], idx[N], out[N], ridx[N];
namespace hld {
int sz[N], son[N], top[N], clk;
void predfs(int u, int d) {
dep[u] = d; sz[u] = 1;
int& maxs = son[u] = -1;
for (int& v: G[u]) {
if (v == fa[u]) continue;
fa[v] = u;
predfs(v, d + 1);
sz[u] += sz[v];
if (maxs == -1 || sz[v] > sz[maxs]) maxs = v;
}
}
void dfs(int u, int tp) {
top[u] = tp; idx[u] = ++clk; ridx[clk] = u;
if (son[u] != -1) dfs(son[u], tp);
for (int& v: G[u])
if (v != fa[u] && v != son[u]) dfs(v, v);
out[u] = clk;
}
template<typename T>
int go(int u, int v, T&& f = [](int, int) {}) {
int uu = top[u], vv = top[v];
while (uu != vv) {
if (dep[uu] < dep[vv]) { swap(uu, vv); swap(u, v); }
f(idx[uu], idx[u]);
u = fa[uu]; uu = top[u];
}
if (dep[u] < dep[v]) swap(u, v);
f(idx[v], idx[u]);
return v;
}
}

int now;
namespace tree {
#define lson l, (l + r) / 2
#define rson (l + r) / 2 + 1, r
extern struct P *null, *pit, pool[];
struct P {
P *ls, *rs; LL sum; int addv; int time;
P* check() {
if (now == time) return this;
if (pit == pool) assert(0);
P* t = new (pit++) P; *t = *this; t->time = now; return t;
}
P* _do_add(int v, int l, int r) {
addv += v; sum += 1LL * (r - l + 1) * v; return this;
}
P* do_add(int v, int l, int r) {
if (this == null) return this;
return check()->_do_add(v, l, r);
}
P* _down(int l, int r) {
if (addv) {
ls = ls->do_add(addv, lson);
rs = rs->do_add(addv, rson);
addv = 0;
}
return this;
}
P* up() { sum = ls->sum + rs->sum; return this; }
P* down(int l, int r) { return check()->_down(l, r); }
} pool[N * 60], *pit = pool, *null = new P{0, 0, 0, 0, 0};
LL query(P* o, int ql, int qr, int l, int r) {
if (ql > r || l > qr) return 0;
if (ql <= l && r <= qr) return o->sum;
o = o->down(l, r);
return query(o->ls, ql, qr, lson) + query(o->rs, ql, qr, rson);
}
P* update(P* o, int ql, int qr, int v, int l, int r) {
if (ql > r || l > qr) return o;
if (ql <= l && r <= qr) return o->do_add(v, l, r);
else {
o = o->down(l, r);
o->ls = update(o->ls, ql, qr, v, lson);
o->rs = update(o->rs, ql, qr, v, rson);
return o->up();
}
}
template<typename T>
P* build(int l, int r, const T& f) {
P* t = new (pit++) P;
if (l == r) { t->sum = f(l); t->ls = t->rs = null; return t; }
t->ls = build(lson, f); t->rs = build(rson, f);
return t->up();
}
}

int w[N];
tree::P* rt[N * 2], *pp[N * 2];

char s[100];
int main() {
#ifdef zerol
freopen("inin", "r", stdin);
#endif
tree::null->ls = tree::null->rs = tree::null;
int n, m; cin >> n >> m;
FOR (i, 1, n + 1) scanf("%d", &w[i]);
tl::init(n);
FOR (_, 0, m) {
int u, v; scanf("%d%d", &u, &v);
tl::G[u].push_back(v); tl::rG[v].push_back(u);
}
tl::go(1); hld::predfs(1, 1); hld::dfs(1, 1);
rt[now = 0] = tree::build(1, n, [&](int x){ return w[ridx[x]]; });
pp[now] = tree::pit;

int Qn; cin >> Qn;
while (Qn--) {
scanf("%s", s);
if (s[0] == 'R') {
int t; scanf("%d", &t);
now = max(0, now - t);
tree::pit = pp[now];
} else {
++now; rt[now] = rt[now - 1];
int tp; scanf("%d", &tp);
if (s[0] == 'C') {
int u, w; scanf("%d%d", &u, &w);
if (tp == 1)
rt[now] = tree::update(rt[now], idx[u], idx[u], w, 1, n);
else if (tp == 2)
rt[now] = tree::update(rt[now], idx[u], out[u], w, 1, n);
else if (tp == 3)
hld::go(1, u, [&](int l, int r){
rt[now] = tree::update(rt[now], l, r, w, 1, n);
});
else assert(0);
pp[now] = tree::pit;
} else if (s[0] == 'Q') {
auto upup = [&](int u) {
LL ret = 0;
hld::go(u, 1, [&](int l, int r) {
ret += tree::query(rt[now], l, r, 1, n);
});
return ret;
};
if (tp == 3) {
static int a[N];
LL ans = 0; int k; scanf("%d", &k);
FOR (i, 0, k) scanf("%d", &a[i]);
sort(a, a + k, [](const int& x, const int& y) { return idx[x] < idx[y]; });
FOR (i, 0, k) ans += upup(a[i]);
FOR (i, 1, k) ans -= upup(hld::go(a[i], a[i - 1], [](int, int){}));
printf("%lld\n", ans);
} else {
int u; scanf("%d", &u);
if (tp == 1)
printf("%lld\n", tree::query(rt[now], idx[u], out[u], 1, n));
else if (tp == 2)
printf("%lld\n", upup(u));
else assert(0);
}
tree::pit = pp[--now];
} else assert(0);
}
}
}
w