题目:
Description
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。Input
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
Output
对于每个/对应的答案输出一行
题解
模板题
用了15mins打完,调了半小时
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(ll &x){
x=0;char ch;bool flag = false;
while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const ll mod = 51061;
const ll maxn = 210010;
struct Node{
Node *ch[2],*fa;
ll mul,add,w,sum,tag;
ll siz;
void update();
void pushdown();
void set();
}*null;
Node mem[maxn],*it;
inline void init(){
it = mem;null = it++;
null->ch[0] = null->ch[1] = null->fa = null;
null->mul = null->add = null->w = null->sum = 0;
null->tag = null->siz = 0;
}
inline Node* newNode(ll w){
Node *p = it++;p->ch[0] = p->ch[1] = p->fa = null;
p->mul = 1;p->siz = 1;
p->add = p->tag = 0;p->w = p->sum = w;return p;
}
inline void Node::update(){
if(this == null) return;
sum = (ch[0]->sum + ch[1]->sum + w) % mod;
siz = (ch[0]->siz + ch[1]->siz + 1);
}
inline void Node::pushdown(){
if(this == null) return;
if(mul != 1){
if(ch[0] != null){
ch[0]->add = ch[0]->add*mul % mod;
ch[0]->mul = ch[0]->mul*mul % mod;
ch[0]->w = ch[0]->w*mul % mod;
ch[0]->sum = ch[0]->sum*mul % mod;
}
if(ch[1] != null){
ch[1]->add = ch[1]->add*mul % mod;
ch[1]->mul = ch[1]->mul*mul % mod;
ch[1]->w = ch[1]->w*mul % mod;
ch[1]->sum = ch[1]->sum*mul % mod;
}
mul = 1;
}
if(add != 0){
if(ch[0] != null){
ch[0]->add = (ch[0]->add + add) % mod;
ch[0]->sum = (ch[0]->sum + add*ch[0]->siz) % mod;
ch[0]->w = (ch[0]->w + add) % mod;
}
if(ch[1] != null){
ch[1]->add = (ch[1]->add + add) % mod;
ch[1]->sum = (ch[1]->sum + add*ch[1]->siz) % mod;
ch[1]->w = (ch[1]->w + add) % mod;
}
add = 0;
}
if(tag != 0){
if(ch[0] != null) ch[0]->tag ^= 1;
if(ch[1] != null) ch[1]->tag ^= 1;
swap(ch[0],ch[1]);tag = 0;
}
}
inline void rotate(Node *p,Node *x){
ll k = p == x->ch[1];
Node *y = p->ch[k^1],*z = x->fa;
if(z->ch[0] == x) z->ch[0] = p;
if(z->ch[1] == x) z->ch[1] = p;
if(y != null) y->fa = x;
p->fa = z;p->ch[k^1] = x;
x->fa = p;x->ch[k] = y;
x->update();p->update();
}
inline bool isroot(Node *p){
return (p == null) || (p->fa->ch[0] != p && p->fa->ch[1] != p);
}
inline void Splay(Node *p){
p->pushdown();
while(!isroot(p)){
Node *x = p->fa,*y = x->fa;
y->pushdown();x->pushdown();p->pushdown();
if(isroot(x)) rotate(p,x);
else if((p == x->ch[0])^(x == y->ch[0])) rotate(p,x),rotate(p,y);
else rotate(x,y),rotate(p,x);
}p->update();
}
inline Node* Access(Node *x){
for(Node *y = null;x != null;y = x,x = x->fa)
Splay(x),x->ch[1] = y,x->update();
return x;
}
inline void makeRoot(Node *x){
Access(x);Splay(x);x->tag ^= 1;
}
inline void link(Node *x,Node *y){
makeRoot(x);x->fa = y;
}
inline void cut(Node *x,Node *y){
makeRoot(x);Access(y);Splay(y);
y->ch[0] = y->ch[0]->fa = null;
y->update();
}
inline void inc(Node *x,Node *y,ll w){
makeRoot(x);Access(y);Splay(y);
y->add += w;y->add %= mod;
y->sum += w*y->siz;
y->sum %= mod;
y->w += w;y->w %= mod;
}
inline void mul(Node *x,Node *y,ll w){
makeRoot(x);Access(y);Splay(y);
y->add *= w;y->add %= mod;
y->mul *= w;y->mul %= mod;
y->sum *= w;y->sum %= mod;
y->w *= w;y->w %= mod;
}
inline ll query(Node *x,Node *y){
makeRoot(x);Access(y);Splay(y);
return y->sum;
}
int main(){
init();
ll n,q;read(n);read(q);
for(ll i=1;i<=n;++i) newNode(1);
for(ll i=1,u,v;i<n;++i){
read(u);read(v);
link(mem+u,mem+v);
}
char ch;
ll u,v,x;
while(q--){
while(ch = getchar(),ch<'!');
if(ch == '+'){
read(u);read(v);read(x);
inc(mem+u,mem+v,x);
}else if(ch == '-'){
read(u);read(v);
cut(mem+u,mem+v);
read(u);read(v);
link(mem+u,mem+v);
}else if(ch == '*'){
read(u);read(v);read(x);
mul(mem+u,mem+v,x);
}else if(ch == '/'){
read(u);read(v);
printf("%lld
",query(mem+u,mem+v));
}
}
getchar();getchar();
return 0;
}