【数据结构】LinkCutTree
https://oi-wiki.org/ds/lct/
https://www.luogu.com.cn/problem/P1501
和重链剖分的概念有点点像。或者应该叫做实链剖分。
LCT维护的“原树”并非是一棵树,而是一片森林。森林中的树可以切成各种形状的树,也可以把不同的树合并。LCT内部的“辅助树”也不是一棵树,而是一片(更大的)Splay森林。
原树是一棵有根树。支持以下操作:
1、对原树节点x和节点y的链更新某信息
2、对原树节点x和节点y的链查询某信息
3、重新指定原树的根节点
4、删除原树的边x-y
5、添加原树的边x-y,其中y是深度更小的节点
LCT维护一片Splay森林,其中每一棵Splay代表原树的一条深度单调递增/单调递减的链(也就是说不拐弯的链)。每一棵Splay的中序遍历是原树的链顶到链底。所以每个节点的实儿子是唯一的,父亲也是唯一的。
实儿子:原树中,当前节点的儿子,且当前节点处于同一棵Splay中,当前节点知道它的实儿子是谁,实儿子也知道其父亲是谁。
虚儿子:原树中,当前节点的儿子,且当前节点不处于同一棵Splay中,当前节点并不知道它的虚儿子有哪些,但是虚儿子知道其父亲是谁。
struct LinkCutTree {
static const int MAXN = 1e5 + 10;
int ch[MAXN][2], pa[MAXN];
int mul[MAXN], add[MAXN], rev[MAXN];
int val[MAXN], siz[MAXN], sum[MAXN];
inline void Init(int n) {
for(int i = 1; i <= n; ++i) {
ch[i][0] = 0, ch[i][1] = 0, pa[i] = 0;
mul[i] = 1, add[i] = 0, rev[i] = 0;
val[i] = 1, siz[i] = 1, sum[i] = val[i];
}
}
inline void PushUp(int x) {
int t = 0;
siz[x] = 1;
sum[x] = val[x];
if(t = ch[x][0]) {
siz[x] += siz[t];
sum[x] = (sum[x] + sum[t]) % MOD;
}
if(t = ch[x][1]) {
siz[x] += siz[t];
sum[x] = (sum[x] + sum[t]) % MOD;
}
}
inline void PushDown(int x) {
int t = 0;
if(mul[x] != 1) {
if(t = ch[x][0]) {
mul[t] = (1LL * mul[t] * mul[x]) % MOD;
add[t] = (1LL * add[t] * mul[x]) % MOD;
val[t] = (1LL * val[t] * mul[x]) % MOD;
sum[t] = (1LL * sum[t] * mul[x]) % MOD;
}
if(t = ch[x][1]) {
mul[t] = (1LL * mul[t] * mul[x]) % MOD;
add[t] = (1LL * add[t] * mul[x]) % MOD;
val[t] = (1LL * val[t] * mul[x]) % MOD;
sum[t] = (1LL * sum[t] * mul[x]) % MOD;
}
mul[x] = 1;
}
if(add[x]) {
if(t = ch[x][0]) {
add[t] = (add[t] + add[x]) % MOD;
val[t] = (val[t] + add[x]) % MOD;
sum[t] = (sum[t] + 1LL * siz[t] * add[x]) % MOD;
}
if(t = ch[x][1]) {
add[t] = (add[t] + add[x]) % MOD;
val[t] = (val[t] + add[x]) % MOD;
sum[t] = (sum[t] + 1LL * siz[t] * add[x]) % MOD;
}
add[x] = 0;
}
if(rev[x]) {
if(t = ch[x][0]) {
swap(ch[t][0], ch[t][1]);
rev[t] ^= 1;
}
if(t = ch[x][1]) {
swap(ch[t][0], ch[t][1]);
rev[t] ^= 1;
}
rev[x] = 0;
}
}
inline bool Which(int x) {
return pa[x] && ch[pa[x]][1] == x;
}
inline bool NotRoot(int x) {
return pa[x] && (ch[pa[x]][0] == x || ch[pa[x]][1] == x);
}
void Rotate(int x) {
int y = pa[x], z = pa[y], k = Which(x);
if(NotRoot(y))
ch[z][Which(y)] = x;
pa[x] = z;
ch[y][k] = ch[x][!k];
pa[ch[x][!k]] = y;
ch[x][!k] = y;
pa[y] = x;
PushUp(y);
PushUp(x);
}
void PushDownAll(int x) {
if(NotRoot(x))
PushDownAll(pa[x]);
PushDown(x);
}
int Splay(int x) {
PushDownAll(x);
while(NotRoot(x)) {
int y = pa[x];
if(NotRoot(y)) {
if(Which(x) == Which(y))
Rotate(y);
else
Rotate(x);
}
Rotate(x);
}
return x;
}
int Access(int x) {
int y = 0;
while(x) {
Splay(x);
ch[x][1] = y;
PushUp(x);
y = x;
x = pa[x];
}
return y;
}
int MakeRoot(int x) {
Access(x);
Splay(x);
swap(ch[x][0], ch[x][1]);
rev[x] ^= 1;
return x;
}
int FindRoot(int x) {
Access(x);
Splay(x);
while(ch[x][0]) {
PushDown(x);
x = ch[x][0];
}
Splay(x);
return x;
}
int Select(int x, int y) {
MakeRoot(x);
Access(y);
Splay(y);
return y;
}
void Cut(int x, int y) {
Select(x, y);
ch[y][0] = 0;
pa[x] = 0;
}
bool Cut2(int x, int y) {
if(FindRoot(x) == FindRoot(y)) {
Select(x, y);
if(ch[y][0] == x && ch[x][1] == 0) {
ch[y][0] = 0;
pa[x] = 0;
return true;
}
}
return false;
}
void Link(int x, int y) {
MakeRoot(x);
pa[x] = y;
}
bool Link2(int x, int y) {
if(FindRoot(x) != FindRoot(y)) {
Link(x, y);
return true;
}
return false;
}
void Mul(int x, int y, int v) {
Select(x, y);
mul[y] = (1LL * v * mul[y]) % MOD;
add[y] = (1LL * add[y] * v) % MOD;
val[y] = (1LL * val[y] * v) % MOD;
sum[y] = (1LL * sum[y] * v) % MOD;
}
void Add(int x, int y, int v) {
Select(x, y);
add[y] = (add[y] + v) % MOD;
val[y] = (val[y] + v) % MOD;
sum[y] = (sum[y] + 1LL * v * siz[y]) % MOD;
}
ll Sum(int x, int y) {
Select(x, y);
return sum[y];
}
} lct;
Access:把x到原树的根节点的路径变为实路径。
额外提供一个返回值,表示最后一次虚边变实边时,虚边父亲的编号。连续两次Access操作时,第二次Access操作的返回值就是这两个节点的LCA。返回值表示x到根节点所在的链代表的Splay的树根,这个节点已经被旋转到Splay的根,并且其父亲一定为空。
MakeRoot:指定原树的根节点为x(换根)
FindRoot:找x在辅助树的树根,也就是x在原树的实链的链顶
Select:把x和y之间的路径切成一个Splay,根是y。(想要根是谁,最后就Splay一下谁就行)
Cut:切断x和y的边
Cut2:假如x和y之间有边,则切断x和y的边
Link:连接x和y的边
Link2:假如x和y之间不连通,则连接x和y的边
Update:更新点x的权值为v
struct LinkCutTree {
static const int MAXN = 1e5 + 10;
int ch[MAXN][2], pa[MAXN];
int mul[MAXN], add[MAXN], rev[MAXN];
int val[MAXN], siz[MAXN], sum[MAXN];
inline void Init(int n) {
for(int i = 1; i <= n; ++i) {
ch[i][0] = 0, ch[i][1] = 0, pa[i] = 0;
mul[i] = 1, add[i] = 0, rev[i] = 0;
rd(val[i]);
siz[i] = 1, sum[i] = val[i];
}
}
inline void PushUp(int x) {
int t = 0;
siz[x] = 1;
sum[x] = val[x];
if(t = ch[x][0]) {
siz[x] += siz[t];
sum[x] = (sum[x] + sum[t]) % MOD;
}
if(t = ch[x][1]) {
siz[x] += siz[t];
sum[x] = (sum[x] + sum[t]) % MOD;
}
}
inline void PushDown(int x) {
int t = 0;
if(mul[x] != 1) {
if(t = ch[x][0]) {
mul[t] = (1LL * mul[t] * mul[x]) % MOD;
add[t] = (1LL * add[t] * mul[x]) % MOD;
val[t] = (1LL * val[t] * mul[x]) % MOD;
sum[t] = (1LL * sum[t] * mul[x]) % MOD;
}
if(t = ch[x][1]) {
mul[t] = (1LL * mul[t] * mul[x]) % MOD;
add[t] = (1LL * add[t] * mul[x]) % MOD;
val[t] = (1LL * val[t] * mul[x]) % MOD;
sum[t] = (1LL * sum[t] * mul[x]) % MOD;
}
mul[x] = 1;
}
if(add[x]) {
if(t = ch[x][0]) {
add[t] = (add[t] + add[x]) % MOD;
val[t] = (val[t] + add[x]) % MOD;
sum[t] = (sum[t] + 1LL * siz[t] * add[x]) % MOD;
}
if(t = ch[x][1]) {
add[t] = (add[t] + add[x]) % MOD;
val[t] = (val[t] + add[x]) % MOD;
sum[t] = (sum[t] + 1LL * siz[t] * add[x]) % MOD;
}
add[x] = 0;
}
if(rev[x]) {
if(t = ch[x][0]) {
swap(ch[t][0], ch[t][1]);
rev[t] ^= 1;
}
if(t = ch[x][1]) {
swap(ch[t][0], ch[t][1]);
rev[t] ^= 1;
}
rev[x] = 0;
}
}
inline bool Which(int x) {
return pa[x] && ch[pa[x]][1] == x;
}
inline bool NotRoot(int x) {
return pa[x] && (ch[pa[x]][0] == x || ch[pa[x]][1] == x);
}
void Rotate(int x) {
int y = pa[x], z = pa[y], k = Which(x);
if(NotRoot(y))
ch[z][Which(y)] = x;
pa[x] = z;
ch[y][k] = ch[x][!k];
pa[ch[x][!k]] = y;
ch[x][!k] = y;
pa[y] = x;
PushUp(y);
PushUp(x);
}
void PushDownAll(int x) {
if(NotRoot(x))
PushDownAll(pa[x]);
PushDown(x);
}
int Splay(int x) {
PushDownAll(x);
while(NotRoot(x)) {
int y = pa[x];
if(NotRoot(y)) {
if(Which(x) == Which(y))
Rotate(y);
else
Rotate(x);
}
Rotate(x);
}
return x;
}
int Access(int x) {
int y = 0;
while(x) {
Splay(x);
ch[x][1] = y;
PushUp(x);
y = x;
x = pa[x];
}
return y;
}
int MakeRoot(int x) {
Access(x);
Splay(x);
swap(ch[x][0], ch[x][1]);
rev[x] ^= 1;
return x;
}
int FindRoot(int x) {
Access(x);
Splay(x);
while(ch[x][0]) {
PushDown(x);
x = ch[x][0];
}
Splay(x);
return x;
}
int Select(int x, int y) {
MakeRoot(x);
Access(y);
Splay(y);
return y;
}
void Cut(int x, int y) {
Select(x, y);
ch[y][0] = 0;
pa[x] = 0;
}
bool Cut2(int x, int y) {
if(FindRoot(x) == FindRoot(y)) {
Select(x, y);
if(ch[y][0] == x && ch[x][1] == 0) {
ch[y][0] = 0;
pa[x] = 0;
return true;
}
}
return false;
}
void Link(int x, int y) {
MakeRoot(x);
pa[x] = y;
}
bool Link2(int x, int y) {
if(FindRoot(x) != FindRoot(y)) {
Link(x, y);
return true;
}
return false;
}
void Mul(int x, int y, int v) {
Select(x, y);
mul[y] = (1LL * v * mul[y]) % MOD;
add[y] = (1LL * add[y] * v) % MOD;
val[y] = (1LL * val[y] * v) % MOD;
sum[y] = (1LL * sum[y] * v) % MOD;
}
void Add(int x, int y, int v) {
Select(x, y);
add[y] = (add[y] + v) % MOD;
val[y] = (val[y] + v) % MOD;
sum[y] = (sum[y] + 1LL * v * siz[y]) % MOD;
}
ll Sum(int x, int y) {
Select(x, y);
return sum[y];
}
} lct;
维护边的信息
由于LCT中根是会变的,所以不能直接把信息存在点上,要对每条边进行拆边。
维护子树的信息
需要额外统计虚儿子的贡献,由于在Splay中看不到原树的虚儿子,所以LCT对此的维护比较麻烦。
并且要求具备逆元,可以说LCT并不擅长维护子树的信息。