嘟嘟嘟
好题，好题
刚开始突发奇想写了一个(O(n ^ 2))暴力，结果竟然过了？！后来才知道是上传题的人把单个数据点开成了10s……
不过不得不说我这暴力写的挺好看的。删边模仿链表删边，加边的时候遍历其中一棵树，使两棵树染上相同的颜色，这样判联通就能达到(O(1))了。
所以我决定先放一个暴力代码
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 5e4 + 5;
const int maxe = 2e5 + 5;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, m, a[maxn];
struct Edge
{
int nxt, to;
}e[maxe];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
e[++ecnt] = (Edge){head[x], y};
head[x] = ecnt;
}
In ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
int fa[maxn], dep[maxn], col[maxn], Col = 0;
In void dfs(int now, int _f, int Col)
{
fa[now] = _f, dep[now] = dep[_f] + 1;
col[now] = Col;
for(int i = head[now], v; ~i; i = e[i].nxt)
{
if((v = e[i].to) == _f) continue;
dfs(v, now, Col);
}
}
In bool cut(int x, int y)
{
for(int i = head[x], v, pre = 0; ~i; pre = i, i = e[i].nxt)
{
if((v = e[i].to) == y)
{
if(!pre) head[x] = e[i].nxt;
else e[pre].nxt = e[i].nxt;
return 1;
}
}
return 0;
}
In void Cut(int x, int y)
{
if(!cut(x, y)) return;
cut(y, x);
dfs(y, 0, ++Col);
}
In void Link(int x, int y)
{
if(col[x] == col[y]) return;
addEdge(x, y), addEdge(y, x);
dfs(y, x, col[x]);
}
In void add(int x, int y, int d)
{
if(col[x] ^ col[y]) return;
while(x ^ y)
{
if(dep[x] < dep[y]) swap(x, y);
a[x] += d; x = fa[x];
}
a[x] += d;
}
int l[maxn], r[maxn], cnt1 = 0, cnt2 = 0;
In void query(int x, int y)
{
if(col[x] ^ col[y]) {puts("-1"); return;}
cnt1 = cnt2 = 0;
while(x ^ y)
{
if(dep[x] >= dep[y]) l[++cnt1] = x, x = fa[x];
else r[++cnt2] = y, y = fa[y];
}
l[++cnt1] = x;
ll ret1 = 0, ret2 = (1LL * cnt1 + cnt2) * (cnt1 + cnt2 + 1) / 2;
for(int i = 1; i <= cnt1; ++i) ret1 += 1LL * a[l[i]] * i * (cnt1 + cnt2 - i + 1);
for(int i = 1; i <= cnt2; ++i) ret1 += 1LL * a[r[i]] * i * (cnt1 + cnt2 - i + 1);
ll Gcd = gcd(ret1, ret2);
write(ret1 / Gcd), putchar('/'), write(ret2 / Gcd), enter;
}
int main()
{
Mem(head, -1);
n = read(), m = read();
for(int i = 1; i <= n; ++i) a[i] = read();
for(int i = 1; i < n; ++i)
{
int x = read(), y = read();
addEdge(x, y), addEdge(y, x);
}
dfs(1, 0, ++Col);
for(int i = 1; i <= m; ++i)
{
int op = read(), x = read(), y = read();
if(op == 1) Cut(x, y);
else if(op == 2) Link(x, y);
else if(op == 3)
{
int d = read();
add(x, y, d);
}
else query(x, y);
}
return 0;
}
好，那么我们切入正题。 不对，我先吐槽一下：第2和第4两个点我的LCT跑的比暴力还慢，然后别的点和暴力差不多，结果总时间竟然比暴力还慢……调了半天也不知道为啥，我这可活什么劲。
好，那现在真的切入正题了。 有时候维护LCT跟线段树差不多，比如这道题，核心就是pushdown和pushup怎么写。 线段树是连个子区间合并，那么这个LCT就是两条链首尾相连合并成一条链。
关于pushup，我实在写不动了，就扔出一篇博客：[城市旅行题解](https://www.luogu.org/blog/user25308/solution-p4842) 思路就是算出左子树的答案$ans_l$，左子树在整棵树中的贡献$w_l$，右子树同理，那么整棵树的答案就是$w_l + w_r = ans_l + Delta x_l + ans_r + Delta x_r$，其中两个$Delta$是能手算出来的。
关于pushdown，除了期望，我和那篇题解都一样。因为我数学没那位老哥那么巨，小学也没学过奥数，推了一阵子搞出个这么个东西：$d * (frac{n ^ 3 + 2n ^ 2 + n}{2} - sum i ^ 2)$。 然后发现没办法$O(1)$求…… 你以为我就去抄题解了吗？那不可能，别忘了，咱这是信竞，不是数竞，后面那个$sum$直接预处理出来不就完了嘛。
对了，子树大小可能在运算的时候会爆int，别忘强制转换成long long。 ```c++ #include
int n, m;
ll SUM[maxn];
struct Tree
{
int ch[2], fa, siz, rev;
ll val, lzy, sum, lsum, rsum, ans;
}t[maxn];
define ls t[now].ch[0]
define rs t[now].ch[1]
define S t[now].siz
In ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
In void c_rev(int now)
{
swap(ls, rs); swap(t[now].lsum, t[now].rsum);
t[now].rev ^= 1;
}
In void c_add(int now, ll d)
{
t[now].lzy += d; t[now].val += d;
t[now].sum += d * S;
t[now].lsum += ((d * S * (S + 1)) >> 1);
t[now].rsum += ((d * S * (S + 1)) >> 1);
t[now].ans += d * (((1LL * S * S * S + 1LL * S * (S << 1) + S) >> 1) - SUM[S]);
}
In void pushdown(int now)
{
if(t[now].rev)
{
if(ls) c_rev(ls);
if(rs) c_rev(rs);
t[now].rev = 0;
}
if(t[now].lzy)
{
if(ls) c_add(ls, t[now].lzy);
if(rs) c_add(rs, t[now].lzy);
t[now].lzy = 0;
}
}
In void pushup(int now)
{
t[now].siz = t[ls].siz + t[rs].siz + 1;
t[now].sum = t[ls].sum + t[rs].sum + t[now].val;
t[now].lsum = t[ls].lsum + t[rs].lsum + (t[rs].sum + t[now].val) * (t[ls].siz + 1);
t[now].rsum = t[rs].rsum + t[ls].rsum + (t[ls].sum + t[now].val) * (t[rs].siz + 1);
t[now].ans = t[ls].ans + t[rs].ans + t[ls].lsum * (t[rs].siz + 1) + t[rs].rsum * (t[ls].siz + 1) + t[now].val * (t[ls].siz + 1) * (t[rs].siz + 1);
}
In bool n_root(int x)
{
return t[t[x].fa].ch[0] == x || t[t[x].fa].ch[1] == x;
}
In void rotate(int x)
{
int y = t[x].fa, z = t[y].fa, k = (t[y].ch[1] == x);
if(n_root(y)) t[z].ch[t[z].ch[1] == y] = x; t[x].fa = z;
t[y].ch[k] = t[x].ch[k ^ 1], t[t[y].ch[k]].fa = y;
t[x].ch[k ^ 1] = y, t[y].fa = x;
pushup(y), pushup(x);
}
int st[maxn], top = 0;
In void splay(int x)
{
int y = x; st[top = 1] = y;
while(n_root(y)) st[++top] = y = t[y].fa;
while(top) pushdown(st[top--]);
while(n_root(x))
{
int y = t[x].fa, z = t[y].fa;
if(n_root(y)) rotate(((t[y].ch[0] == x) ^ (t[z].ch[0] == y)) ? x : y);
rotate(x);
}
}
In void access(int x)
{
int y = 0;
while(x)
{
splay(x); t[x].ch[1] = y;
pushup(x);
y = x; x = t[x].fa;
}
}
In void make_root(int x)
{
access(x), splay(x);
c_rev(x);
}
In int find_root(int x)
{
access(x), splay(x);
while(t[x].ch[0]) pushdown(x), x = t[x].ch[0];
return x;
}
In void split(int x, int y)
{
make_root(x);
access(y), splay(y);
}
In void Link(int x, int y)
{
make_root(x);
if(find_root(y) ^ x) t[x].fa = y;
}
In void Cut(int x, int y)
{
make_root(x);
if(find_root(y) == x && t[x].fa == y && !t[x].ch[1])
t[y].ch[0] = t[x].fa = 0, pushup(y);
}
In void update(int x, int y, int d)
{
make_root(x);
if(find_root(y) ^ x) return;
split(x, y); c_add(y, d);
pushup(y);
}
In void query(int x, int y)
{
make_root(x);
if(find_root(y) ^ x) {puts("-1"); return;}
split(x, y);
ll Siz = t[y].siz, tp = (Siz * (Siz + 1)) >> 1, Gcd = gcd(t[y].ans, tp);
write(t[y].ans / Gcd), putchar('/'), write(tp / Gcd), enter;
}
int main()
{
n = read(), m = read();
for(int i = 1; i <= n; ++i)
{
t[i].val = read(), t[i].siz = 1;
t[i].sum = t[i].lsum = t[i].rsum = t[i].ans = t[i].val;
SUM[i] = SUM[i - 1] + i * i;
}
for(int i = 1; i < n; ++i)
{
int x = read(), y = read();
Link(x, y);
}
for(int i = 1; i <= m; ++i)
{
int op = read(), x = read(), y = read();
if(op == 1) Cut(x, y);
else if(op == 2) Link(x, y);
else if(op == 4) query(x, y);
else
{
int d = read();
update(x, y, d);
}
}
return 0;
}