problem
给出一棵有点权的树,支持4种操作。
- 路径加
- 路径乘
- 删除一条边并添加一条边,操作后还是一棵树
- 求路径和
solution
删边和添边的操作可以用LCT解决。
路径加和路径乘再splay上打标记即可。
下方的时候要按照一定的顺序下方。下面的代码是先下放乘法标记,下放乘法标记的时候需要将加法标记一起乘。
code
/*
* @Author: wxyww
* @Date: 2020-02-25 17:55:05
* @Last Modified time: 2020-02-25 20:22:06
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned int uin;
const int N = 100010,mod = 51061;
#define ls TR[cur].ch[0]
#define rs TR[cur].ch[1]
ll read() {
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9') {
if(c=='-') f=-1;
c=getchar();
}
while(c>='0'&&c<='9') {
x=x*10+c-'0';
c=getchar();
}
return x*f;
}
struct node {
uin pre,ch[2],val,sum,lazy1,lazy2,siz,rev;
}TR[N];
void up(int cur) {
TR[cur].sum = (TR[ls].sum + TR[rs].sum + TR[cur].val) % mod;
TR[cur].siz = TR[ls].siz + TR[rs].siz + 1;
}
void pushdown(int cur) {
if(TR[cur].lazy1 != 1) {//lazy1是乘法标记
uin tmp = TR[cur].lazy1;
TR[ls].lazy1 = TR[ls].lazy1 * tmp % mod;
TR[rs].lazy1 = TR[rs].lazy1 * tmp % mod;
TR[ls].lazy2 = TR[ls].lazy2 * tmp % mod;
TR[rs].lazy2 = TR[rs].lazy2 * tmp % mod;
TR[ls].sum = TR[ls].sum * tmp % mod;
TR[rs].sum = TR[rs].sum * tmp % mod;
TR[ls].val = TR[ls].val * tmp % mod;
TR[rs].val = TR[rs].val * tmp % mod;
TR[cur].lazy1 = 1;
}
if(TR[cur].lazy2) {//lazy2为加法标记
uin tmp = TR[cur].lazy2;
TR[ls].lazy2 = (TR[ls].lazy2 + tmp) % mod;
TR[rs].lazy2 = (TR[rs].lazy2 + tmp) % mod;
TR[ls].sum = (TR[ls].sum + tmp * TR[ls].siz % mod) % mod;
TR[rs].sum = (TR[rs].sum + tmp * TR[rs].siz % mod) % mod;
TR[ls].val = (TR[ls].val + tmp) % mod;
TR[rs].val = (TR[rs].val + tmp) % mod;
TR[cur].lazy2 = 0;
}
if(TR[cur].rev) {
TR[ls].rev ^= 1;TR[rs].rev ^= 1;
swap(ls,rs);
TR[cur].rev = 0;
}
}
int isroot(int cur) {
return TR[TR[cur].pre].ch[0] != cur && TR[TR[cur].pre].ch[1] != cur;
}
int getwh(int cur) {
return TR[TR[cur].pre].ch[1] == cur;
}
void rotate(int cur) {
int fa = TR[cur].pre,gr = TR[fa].pre,f = getwh(cur);
if(!isroot(fa)) TR[gr].ch[getwh(fa)] = cur;
TR[cur].pre = gr;
if(TR[cur].ch[f ^ 1]) TR[TR[cur].ch[f ^ 1]].pre = fa;
TR[fa].ch[f] = TR[cur].ch[f ^ 1];
TR[fa].pre = cur;TR[cur].ch[f ^ 1] = fa;
up(fa);up(cur);
}
int sta[N],top;
void splay(int cur) {
sta[++top] = cur;
for(int i = cur;!isroot(i);i = TR[i].pre) sta[++top] = TR[i].pre;
while(sta[top]) pushdown(sta[top--]);
while(!isroot(cur)) {
if(!isroot(TR[cur].pre)) {
if(getwh(TR[cur].pre) == getwh(cur)) rotate(TR[cur].pre);
else rotate(cur);
}
rotate(cur);
}
}
void access(int cur) {
for(int t = 0;cur;t = cur,cur = TR[cur].pre) {
splay(cur);rs = t;up(cur);
}
}
void makeroot(int cur) {
access(cur);splay(cur);
TR[cur].rev ^= 1;
}
void link(int x,int y) {
makeroot(x);TR[x].pre = y;
}
void cut(int x,int cur) {
makeroot(x);access(cur);
splay(cur);
ls = TR[ls].pre = 0;
up(cur);
}
void update2(int x,int y,int c) {//处理加法
makeroot(x);access(y);splay(y);
TR[y].sum += c * TR[y].siz % mod;
TR[y].sum %= mod;
TR[y].val = (TR[y].val + c) % mod;
TR[y].lazy2 = (TR[y].lazy2 + c) % mod;
}
void update1(int x,int y,int c) {
makeroot(x);access(y);splay(y);
TR[y].sum = TR[y].sum * c % mod;
TR[y].lazy1 = TR[y].lazy1 * c % mod;
TR[y].val = TR[y].val * c % mod;
TR[y].lazy2 = TR[y].lazy2 * c % mod;
}
int query(int x,int y) {
makeroot(x);access(y);splay(y);
return TR[y].sum;
}
signed main() {
int n = read(),Q = read();
for(int i = 1;i <= n;++i) TR[i].lazy1 = TR[i].val = TR[i].sum = TR[i].siz = 1;
for(int i = 1;i < n;++i) {
int u = read(),v = read();
link(u,v);
}
while(Q--) {
char c = getchar();
while(c != '+' && c != '-' && c != '*' && c != '/') c = getchar();
if(c == '+') {
int u = read(),v = read(),w = read();
update2(u,v,w);
}
if(c == '-') {
int u1 = read(),v1 = read(),u2 = read(),v2 = read();
cut(u1,v1);link(u2,v2);
}
if(c == '*') {
int u = read(),v = read(),w = read();
update1(u,v,w);
}
if(c == '/') {
int u = read(),v = read();
printf("%d
",query(u,v));
}
}
return 0;
}