luogu1501 [国家集训队]Tree II

题目链接

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\n",query(u,v));
		}
	}

	return 0;
}