题目
一棵n个点的树,每个点的初始权值为1。对于这棵树有q个操作,每个操作为以下四种操作之一:
+ u v c:将u到v的路径上的点的权值都加上自然数c;
- u1 v1 u2 v2:将树中原有的边(u1,v1)删除,加入一条新边(u2,v2),保证操作完之后仍然是一棵树;
* u v c:将u到v的路径上的点的权值都乘上自然数c;
/ u v:询问u到v的路径上的点的权值和,求出答案对于51061的余数。
输入格式
第一行两个整数n,q
接下来n-1行每行两个正整数u,v,描述这棵树
接下来q行,每行描述一个操作
输出格式
对于每个/对应的答案输出一行
输入样例
3 2
1 2
2 3
- 1 3 4
/ 1 1
输出样例
4
题解
好久没敲LCT了。都快忘了【本来也没敲过几次】
一道很裸的板题,只是标记多了些,不过也是很经典的乘法 + 加法标记
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)
#define isrt(u) (!e[u].f || (e[e[u].f].ch[0] != u && e[e[u].f].ch[1] != u))
#define isr(u) (e[e[u].f].ch[1] == u)
#define ls e[u].ch[0]
#define rs e[u].ch[1]
using namespace std;
const int maxn = 100005,maxm = 200005,INF = 1000000000,P = 51061;
inline int RD(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();}
return out * flag;
}
int h[maxn],ne = 0,N,Q;
struct EDGE{int to,nxt;}ed[maxm];
struct node{LL ch[2],f,plu,tim,w,sum,rev,siz;}e[maxn];
inline void build(int u,int v){
ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;
ed[ne] = (EDGE){u,h[v]}; h[v] = ne++;
}
void pup(int u){
e[u].sum = (e[u].w + e[ls].sum + e[rs].sum) % P;
e[u].siz = e[ls].siz + e[rs].siz + 1;
}
void pd(int u){
if (e[u].tim != 1){
e[ls].w = e[ls].w * e[u].tim % P;
e[rs].w = e[rs].w * e[u].tim % P;
e[ls].sum = e[ls].sum * e[u].tim % P;
e[rs].sum = e[rs].sum * e[u].tim % P;
e[ls].plu = e[ls].plu * e[u].tim % P;
e[rs].plu = e[rs].plu * e[u].tim % P;
e[ls].tim = e[ls].tim * e[u].tim % P;
e[rs].tim = e[rs].tim * e[u].tim % P;
}
if (e[u].plu){
e[ls].w = (e[ls].w + e[u].plu) % P;
e[rs].w = (e[rs].w + e[u].plu) % P;
e[ls].sum = (e[ls].sum + (LL)e[ls].siz * e[u].plu % P) % P;
e[rs].sum = (e[rs].sum + (LL)e[rs].siz * e[u].plu % P) % P;
e[ls].plu = (e[ls].plu + e[u].plu) % P;
e[rs].plu = (e[rs].plu + e[u].plu) % P;
}
if (e[u].rev){
swap(ls,rs);
e[ls].rev ^= 1;
e[rs].rev ^= 1;
}
e[u].plu = e[u].rev = 0; e[u].tim = 1;
}
void push_down(int u){if (!isrt(u)) push_down(e[u].f); pd(u);}
void spin(int u){
int s = isr(u),fa = e[u].f;
e[u].f = e[fa].f;
if (!isrt(fa)) e[e[fa].f].ch[isr(fa)] = u;
e[fa].ch[s] = e[u].ch[s ^ 1];
if (e[u].ch[s ^ 1]) e[e[u].ch[s ^ 1]].f = fa;
e[fa].f = u;
e[u].ch[s ^ 1] = fa;
pup(fa);pup(u);
}
void splay(int u){
push_down(u);
while (!isrt(u)){
if (isrt(e[u].f)) spin(u);
else if (isr(u) ^ isr(e[u].f)) spin(u),spin(u);
else spin(e[u].f),spin(u);
}
}
void Access(int u){
int v = 0;
while (u) splay(u),rs = v,e[v].f = u,pup(u),u = e[v = u].f;
}
void Make_root(int u){Access(u); splay(u); e[u].rev ^= 1;}
void Link(int u,int v){
Make_root(u); Access(v); splay(v);
e[u].f = v; pup(v);
}
void Cut(int u,int v){
Make_root(u); Access(v); splay(v);
e[v].ch[0] = e[u].f = 0; pup(v);
}
LL Query(int u,int v){
Make_root(u); Access(v); splay(v);
return e[v].sum;
}
void Add(int u,int v,LL w){
Make_root(u); Access(v); splay(v);
e[v].w = (e[v].w + w) % P;
e[v].plu = (e[v].plu + w) % P;
e[v].sum = (e[v].sum + (LL)w * e[v].siz % P) % P;
}
void Tim(int u,int v,LL w){
Make_root(u); Access(v); splay(v);
e[v].w = (e[v].w * w) % P;
e[v].sum = (e[v].sum * w) % P;
e[v].tim = (e[v].tim * w) % P;
e[v].plu = (e[v].plu * w) % P;
}
void dfs(int u){
Redge(u) if (ed[k].to != e[u].f){
e[ed[k].to].f = u;
dfs(ed[k].to);
}
}
int main(){
memset(h,-1,sizeof(h));
N = RD(); Q = RD(); REP(i,N - 1) build(RD(),RD());
REP(i,N) e[i].sum = e[i].w = e[i].tim = e[i].siz = 1;
dfs(1); char opt; int u,v,u1,v1;
while (Q--){
opt = getchar();
while (opt != '+' && opt != '-' && opt != '*' && opt != '/') opt = getchar();
u = RD(); v = RD();
if (opt == '+') Add(u,v,RD());
else if (opt == '-') u1 = RD(),v1 = RD(),Cut(u,v),Link(u1,v1);
else if (opt == '*') Tim(u,v,RD());
else if (opt == '/') printf("%lld
",Query(u,v));
}
return 0;
}