P1501 [国家集训队]Tree II(LCT 下推标记,LCT 维护树上信息)
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2022-04-06 23:48:31
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用 LCT 维护整棵树,splay 中要维护每个点的权值,子树节点个数以及子树和,为了优化复杂度还要维护下推标记。
这题有三种标记:翻转,加法,乘法
翻转标记的下推顺序不影响维护值,加法和乘法优先维护乘法下推标记,将乘法下推时要将子节点的子树和,节点权值,加法下推标记和乘法下推标记全部乘上乘法标记
加法下推标记则直接加即可,子树和要加上加法下推标记 * 子树节点个数
代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
typedef long long ll;
int n,m;
const int mod = 51061;
inline int read(){
int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}
struct LCT { //用splay维护原森林的连通,用到了splay的操作以及数组
int ch[maxn][2]; //ch[u][0] 表示 左二子,ch[u][1] 表示右儿子
int f[maxn]; //当前节点的父节点
int tag[maxn]; //翻转标记,乘标记,加标记
int top,sta[maxn],sz[maxn];
ll sum[maxn],val[maxn],add[maxn],mul[maxn];
inline bool get(int x) {
return ch[f[x]][1] == x;
}
void init() {
memset(f,0,sizeof f);
memset(ch,0,sizeof ch);
memset(tag,0,sizeof tag);
memset(add,0,sizeof add);
for (int i = 1; i <= maxn - 10; i++)
sum[i] = val[i] = mul[i] = sz[i] = 1;
}
inline void pushup(int rt) {
if (rt) {
sz[rt] = 1; sum[rt] = val[rt];
if (ch[rt][0]) sz[rt] += sz[ch[rt][0]], sum[rt] = (sum[rt] + sum[ch[rt][0]]) % mod;
if (ch[rt][1]) sz[rt] += sz[ch[rt][1]], sum[rt] = (sum[rt] + sum[ch[rt][1]]) % mod;
}
}
inline void pushdown(int rt) {
if (tag[rt]) {
int ls = ch[rt][0], rs = ch[rt][1];
if (ls) swap(ch[ls][0],ch[ls][1]), tag[ls] ^= 1;
if (rs) swap(ch[rs][0],ch[rs][1]), tag[rs] ^= 1;
tag[rt] = 0;
}
if (mul[rt] != 1) {
if (ch[rt][0]) {
val[ch[rt][0]] = val[ch[rt][0]] * mul[rt] % mod;
add[ch[rt][0]] = add[ch[rt][0]] * mul[rt] % mod;
mul[ch[rt][0]] = mul[ch[rt][0]] * mul[rt] % mod;
sum[ch[rt][0]] = sum[ch[rt][0]] * mul[rt] % mod;
}
if (ch[rt][1]) {
val[ch[rt][1]] = val[ch[rt][1]] * mul[rt] % mod;
add[ch[rt][1]] = add[ch[rt][1]] * mul[rt] % mod;
mul[ch[rt][1]] = mul[ch[rt][1]] * mul[rt] % mod;
sum[ch[rt][1]] = sum[ch[rt][1]] * mul[rt] % mod;
}
mul[rt] = 1;
}
if (add[rt]) {
if (ch[rt][0]) {
val[ch[rt][0]] = (val[ch[rt][0]] + add[rt]) % mod;
add[ch[rt][0]] = (add[ch[rt][0]] + add[rt]) % mod;
sum[ch[rt][0]] = (sum[ch[rt][0]] + 1ll * add[rt] * sz[ch[rt][0]]) % mod;
}
if (ch[rt][1]) {
val[ch[rt][1]] = (val[ch[rt][1]] + add[rt]) % mod;
add[ch[rt][1]] = (add[ch[rt][1]] + add[rt]) % mod;
sum[ch[rt][1]] = (sum[ch[rt][1]] + 1ll * add[rt] * sz[ch[rt][1]]) % mod;
}
add[rt] = 0;
}
}
inline bool isroot(int x) {
return (ch[f[x]][0] != x) && (ch[f[x]][1] != x);
}
inline void rotate(int x) { //旋转操作,根据 x 在 f[x] 的哪一侧进行左旋和右旋
int old = f[x], oldf = f[old];
int whichx = get(x);
if(!isroot(old)) ch[oldf][ch[oldf][1] == old] = x; //如果 old 不是根节点,就要修改 oldf 的子节点信息
ch[old][whichx] = ch[x][whichx ^ 1];
ch[x][whichx ^ 1] = old;
f[ch[old][whichx]] = old;
f[old] = x; f[x] = oldf;
pushup(old); pushup(x);
}
inline void splay(int x) { //将 x 旋到所在 splay 的根
top = 0; sta[++top] = x;
for (int i = x; !isroot(i); i = f[i]) sta[++top] = f[i]; //在 splay 中维护 下推标记
while(top) pushdown(sta[top--]);
for(int fa = f[x]; !isroot(x); rotate(x), fa = f[x]) { //再把x翻上来
if(!isroot(fa)) //如果fa非根,且x 和 fa是同一侧,那么先翻转fa,否则先翻转x
rotate((get(x) == get(fa)) ? fa : x);
}
}
inline void access(int x) { //access操作将x 到 根路径上的边修改为重边
int lst = 0;
while(x > 0) {
splay(x);
ch[x][1] = lst;
pushup(x);
lst = x; x = f[x];
}
}
inline void move_to_root(int x) { //将 x 移到 x 所在树的根(不是所在splay的根,所在splay只是一条重链)
access(x); splay(x); tag[x] ^= 1; swap(ch[x][0],ch[x][1]);
//将 x 移到 根之后 x 是深度最低的点,这条重链、这棵splay上所有点的深度颠倒,
//所有的点的左子树的点应该到右子树,因此要翻转这棵splay的左右子树
}
inline int findroot(int x) {
access(x);
splay(x);
int rt = x;
while(ch[rt][0]) rt = ch[rt][0];
return rt;
}
inline void link(int x,int y) {
move_to_root(x); f[x] = y; splay(x);
}
inline void cut(int x,int y) {
move_to_root(x); access(y);
splay(y); ch[y][0] = f[x] = 0;
pushup(y);
}
}tree;
int x,y,u,v,k[maxn];
char op[10];
int main() {
n = read(); m = read();
tree.init();
for (int i = 1; i < n; i++) {
u = read(); v = read();
tree.link(u,v);
}
while(m--) {
scanf("%s",op);
if (op[0] == '+') {
u = read(), v = read(), x = read();
tree.move_to_root(u);
tree.access(v);
tree.splay(v);
tree.val[v] = (tree.val[v] + x) % mod;
tree.add[v] = (tree.add[v] + x) % mod;
tree.sum[v] = (tree.sum[v] + 1ll * x * tree.sz[v]) % mod;
}
if (op[0] == '-') {
u = read(); v = read(); x = read(); y = read();
tree.cut(u,v);
tree.link(x,y);
}
if (op[0] == '*') {
u = read(); v = read(); x = read();
tree.move_to_root(u);
tree.access(v);
tree.splay(v);
tree.val[v] = tree.val[v] * x % mod;
tree.mul[v] = tree.mul[v] * x % mod;
tree.add[v] = tree.add[v] * x % mod;
tree.sum[v] = tree.sum[v] * x % mod;
}
if (op[0] == '/') {
u = read(); v = read();
tree.move_to_root(u);
tree.access(v);
tree.splay(v);
printf("%lld\n",tree.sum[v]);
}
}
return 0;
}