# bzoj3091 城市旅行 LCT + 区间合并

## 题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=3091

## 题解

1. $ls, rs, sum$ 一类东西太多，不要搞混；
2. 不连通时不要操作！！！不连通时不要操作！！！不连通时不要操作！！！

#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;
}

const int N = 50000 + 7;

#define lc c[0]
#define rc c[1]

int n, m;
int a[N];

struct Node {
int c[2], fa, rev, sz, v, add;
ll s, sum, ls, rs;
} t[N];
int st[N];
inline bool isroot(int o) { return t[t[o].fa].lc != o && t[t[o].fa].rc != o; }
inline bool idtfy(int o) { return t[t[o].fa].rc == o; }
inline void connect(int fa, int o, int d) { t[fa].c[d] = o, t[o].fa = fa; }
inline Node add_tag(Node a, int k) {
a.v += k, a.add += k;
a.s += (ll)k * a.sz;
a.sum += (ll)a.sz * (a.sz + 1) * (a.sz + 2) / 6 * k;
a.ls += (ll)a.sz * (a.sz + 1) / 2 * k;
a.rs += (ll)a.sz * (a.sz + 1) / 2 * k;
return a;
}
inline void pushup(int o) {
const Node &tl = t[t[o].lc], &tr = t[t[o].rc];
t[o].sz = tl.sz + tr.sz + 1;
t[o].s = tl.s + tr.s + t[o].v;
t[o].ls = tl.ls + (tl.s + t[o].v) * (tr.sz + 1) + tr.ls;
t[o].rs = tr.rs + (tr.s + t[o].v) * (tl.sz + 1) + tl.rs;
t[o].sum = tl.sum + tr.sum + (ll)t[o].v * (tl.sz + 1) * (tr.sz + 1) + tl.rs * (tr.sz + 1) + tr.ls * (tl.sz + 1);
}
inline void pushdown(int o) {
if (t[o].rev) {
if (t[o].lc) t[t[o].lc].rev ^= 1, std::swap(t[t[o].lc].lc, t[t[o].lc].rc), std::swap(t[t[o].lc].ls, t[t[o].lc].rs);
if (t[o].rc) t[t[o].rc].rev ^= 1, std::swap(t[t[o].rc].lc, t[t[o].rc].rc), std::swap(t[t[o].rc].ls, t[t[o].rc].rs);
t[o].rev = 0;
}
}
}
inline void rotate(int o) {
int fa = t[o].fa, pa = t[fa].fa, d1 = idtfy(o), d2 = idtfy(fa), b = t[o].c[d1 ^ 1];
if (!isroot(fa)) t[pa].c[d2] = o; t[o].fa = pa;
connect(o, fa, d1 ^ 1), connect(fa, b, d1);
pushup(fa), pushup(o);
}
inline void splay(int o) {
int x = o, tp = 0;
st[++tp] = x;
while (!isroot(x)) st[++tp] = x = t[x].fa;
while (tp) pushdown(st[tp--]);
while (!isroot(o)) {
int fa = t[o].fa;
if (isroot(fa)) rotate(o);
else if (idtfy(o) == idtfy(fa)) rotate(fa), rotate(o);
else rotate(o), rotate(o);
}
}
inline void access(int o) {
for (int x = 0; o; o = t[x = o].fa)
splay(o), t[o].rc = x, pushup(o);
}
inline void mkrt(int o) {
access(o), splay(o);
t[o].rev ^= 1, std::swap(t[o].lc, t[o].rc), std::swap(t[o].ls, t[o].rs);
}
inline int getrt(int o) {
access(o), splay(o);
while (pushdown(o), t[o].lc) o = t[o].lc;
return splay(o), o;
}
inline void link(int x, int y) {
mkrt(x);
if (getrt(y) != x) t[x].fa = y;
}
inline void cut(int x, int y) {
mkrt(x), access(y), splay(y);
if (t[y].lc == x && !t[x].rc) t[y].lc = t[x].fa = 0, pushup(y);
}

inline void work() {
while (m--) {
int opt, x, y, d;
if (opt == 1) cut(x, y);
else if (opt == 2) link(x, y);
else if (opt == 3) read(d), getrt(x) == getrt(y) && (mkrt(x), access(y), splay(y), t[y] = add_tag(t[y], d), 1);
else if (getrt(x) != getrt(y)) puts("-1");
else {
mkrt(x), access(y), splay(y);
ll sum = t[y].sum, cnt = (ll)t[y].sz * (t[y].sz + 1) / 2, p = std::__gcd(sum, cnt);
printf("%lld/%lld\n", sum / p, cnt / p);
}
}
}

inline void init() {
for (int i = 1; i <= n; ++i) read(a[i]);
for (int i = 1; i <= n; ++i) t[i].v = a[i], pushup(i);
int x, y;
for (int i = 1; i < n; ++i) read(x), read(y), link(x, y);
}

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

posted @ 2019-10-22 22:27  hankeke  阅读(129)  评论(0编辑  收藏  举报