调了一下午。。
属实自闭
想法很简单
同Sdoi2016游戏的做法
将路径拆成两条 \(u \to lca\)，\(lca \to v\)
等差数列对 \(u \to lca\) 路径上的点 \(x\) 的影响为 \(a + (dep_{u}-dep_{x})\times b\)
可以拆成 \((-b) \times dep_x + (a + dep_{u} \times b)\)
变成了 \(kx + b\) 的形式
给每个区间维护一个 \(pair\) \((k,b)\) 就很好快速求和了
\(lca \to v\) 同理，注意 \(lca\) 处不要加两次
区间修改就标记永久化解决
注意版本回退的问题
以前喜欢给每个版本编号，然后通过编号回退版本，比如回退到版本 \(x\)
但是这样下一次修改就会对 \(x+1\) 个版本进行修改。这样是错的。（原来之前一直是错的。。）
直接用一个 \(rt\) 变量回退到 \(root_x\) 处，再新建版本才是对的。
#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp tree[p].l
#define rp tree[p].r
#define mid ((l + r) / 2)
#define lowbit(i) ((i) & (-i))
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
int x = 0, f = 1; char ch = getc();
while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
return x * f;
}
inline char cc() {
char ch = getc();
while (!(ch >= 'a' && ch <= 'z')) ch = getc();
return ch;
}
inline ll __() {
ll x = 0, f = 1; char ch = getc();
while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
return x * f;
}
const int N = 1e5 + 7, E = 2e5 + 7;
Edg
int dep[N], son[N], fa[N], sz[N], n, m, in[N], dfn, mp[N], top[N];
ll dis[N];
void dfs1(int u, int f = 0) {
dep[u] = dep[f] + 1; fa[u] = f;
sz[u] = 1;
es (u, i, v) {
if (v == f) continue;
dfs1(v, u);
sz[u] += sz[v];
if (sz[son[u]] < sz[v]) son[u] = v;
}
}
void dfs2(int u, int tp) {
top[u] = tp; in[u] = ++dfn; mp[dfn] = u;
if (!son[u]) return;
dfs2(son[u], tp);
es (u, i, v) {
if (v != fa[u] && v != son[u]) dfs2(v, v);
}
}
int Lca(int u, int v) {
while (top[u] != top[v]) {
if (dep[top[u]] < dep[top[v]]) std::swap(u, v);
u = fa[top[u]];
}
if (dep[u] > dep[v]) std::swap(u, v);
return u;
}
struct Node {
ll first, second;
Node() {}
Node(ll f, ll s): first(f), second(s) {}
};
struct Seg {
struct Tree {
int l, r;
ll k, b;
ll sum;
} tree[N * 90];
int tol;
void update(int &p, int l, int r, int x, int y, const Node &li) {
int p0 = p;
p = ++tol;
tree[p] = tree[p0];
int L = std::max(l, x), R = std::min(r, y);
tree[p].sum += (ll)(dis[R] - dis[L - 1]) * li.fi + (ll)(R - L + 1) * li.se;
if (x <= l && y >= r) {
tree[p].k += li.fi; tree[p].b += li.se;
return;
}
if (x <= mid) update(lp, l, mid, x, y, li);
if (y > mid) update(rp, mid + 1, r, x, y, li);
}
ll query(int p, int l, int r, int x, int y) {
if (!p || (x <= l && y >= r)) return tree[p].sum;
int L = std::max(l, x), R = std::min(r, y);
ll ans = (ll)(dis[R] - dis[L - 1]) * tree[p].k + (ll)(R - L + 1) * tree[p].b;
if (x <= mid) ans += query(lp, l, mid, x, y);
if (y > mid) ans += query(rp, mid + 1, r, x, y);
return ans;
}
} seg;
int root[N], cur, rt;
void modify(int u, int v, ll a, ll b) {
int lca = Lca(u, v);
Node o(-a, b + dep[u] * a);
while (top[u] != top[lca]) {
seg.update(rt, 1, n, in[top[u]], in[u], o);
u = fa[top[u]];
}
seg.update(rt, 1, n, in[lca], in[u], o);
o.fi = a, o.se -= 2 * dep[lca] * a;
while (top[v] != top[lca]) {
seg.update(rt, 1, n, in[top[v]], in[v], o);
v = fa[top[v]];
}
if (in[v] == in[lca]) return;
seg.update(rt, 1, n, in[lca] + 1, in[v], o);
}
ll query(int u, int v) {
ll ans = 0;
while (top[u] != top[v]) {
if (dep[top[u]] < dep[top[v]]) std::swap(u, v);
ans += seg.query(rt, 1, n, in[top[u]], in[u]);
u = fa[top[u]];
}
if (dep[u] > dep[v]) std::swap(u, v);
ans += seg.query(rt, 1, n, in[u], in[v]);
return ans;
}
signed main() {
#ifdef LOCAL
freopen("ans.out", "w", stdout);
#endif
n = _(), m = _();
rep (i, 1, n) {
int u, v;
u = _(), v = _();
add(u, v);
}
dfs1(1);
dfs2(1, 1);
rep (i, 1, n + 1) dis[i] = dis[i - 1] + dep[mp[i]];
ll ans = 0;
rt = root[0] = 0;
rep (i, 0, m) {
char opt = cc();
ll x1, y1, a, b;
if (opt == 'c') {
x1 = __(), y1 = __(), a = _(), b = _();
x1 ^= ans; y1 ^= ans;
modify((int)x1, (int)y1, b, a);
root[++cur] = rt;
} else if (opt == 'q') {
x1 = __(), y1 = __();
x1 ^= ans; y1 ^= ans;
printf("%lld\n", ans = query((int)x1, (int)y1));
} else {
x1 = __();
x1 ^= ans;
rt = root[x1];
}
}
#ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
return 0;
}