【bzoj2631】tree LCT

题目描述

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

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

对于每个/对应的答案输出一行

样例输入

3 2
1 2
2 3
* 1 3 4
/ 1 1

样例输出

4


题解

带点权的LCT

需要注意的是3标记的处理:先乘后加,同时乘标记,与翻转互不影响。

这题的坑点在于int会WA,long long会TLE,必须用unsigned int。

#include <cstdio>
#include <algorithm>
#define N 100010
#define MOD 51061
#define lson c[0][x]
#define rson c[1][x]
using namespace std;
int fa[N] , c[2][N] , si[N] , rev[N];
unsigned w[N] , sum[N] , add[N] , mul[N];
char str[5];
inline int read()
{
	int ret = 0; char ch = getchar();
	while(ch < '0' || ch > '9') ch = getchar();
	while(ch >= '0' && ch <= '9') ret = (ret << 3) + (ret << 1) + ch - '0' , ch = getchar();
	return ret;
}
void pushup(int x)
{
	si[x] = si[lson] + si[rson] + 1;
	sum[x] = (sum[lson] + sum[rson] + w[x]) % MOD;
}
void cal(int x , unsigned a , unsigned m , int r)
{
	sum[x] = (sum[x] * m + si[x] * a) % MOD;
	w[x] = (w[x] * m + a) % MOD;
	mul[x] = (mul[x] * m) % MOD;
	add[x] = (add[x] * m + a) % MOD;
	if(r) swap(lson , rson) , rev[x] ^= 1;
}
void pushdown(int x)
{
	cal(lson , add[x] , mul[x] , rev[x]);
	cal(rson , add[x] , mul[x] , rev[x]);
	add[x] = rev[x] = 0 , mul[x] = 1;
}
bool isroot(int x)
{
	return c[0][fa[x]] != x && c[1][fa[x]] != x;
}
void update(int x)
{
	if(!isroot(x)) update(fa[x]);
	pushdown(x);
}
void rotate(int x)
{
	int y = fa[x] , z = fa[y] , l = (c[1][y] == x) , r = l ^ 1;
	if(!isroot(y)) c[c[1][z] == y][z] = x;
	fa[x] = z , fa[y] = x , fa[c[r][x]] = y , c[l][y] = c[r][x] , c[r][x] = y;
	pushup(y) , pushup(x);
}
void splay(int x)
{
	update(x);
	while(!isroot(x))
	{
		int y = fa[x] , z = fa[y];
		if(!isroot(y))
		{
			if((c[0][y] == x) ^ (c[0][z] == y)) rotate(x);
			else rotate(y);
		}
		rotate(x);
	}
}
void access(int x)
{
	int t = 0;
	while(x) splay(x) , rson = t , pushup(x) , t = x , x = fa[x];
}
void makeroot(int x)
{
	access(x) , splay(x);
	swap(lson , rson) , rev[x] ^= 1;
}
void link(int x , int y)
{
	makeroot(x) , fa[x] = y;
}
void cut(int x , int y)
{
	makeroot(x) , access(y) , splay(y) , c[0][y] = fa[x] = 0 , pushup(y);
}
void split(int x , int y)
{
	makeroot(y) , access(x) , splay(x);
}
int main()
{
	int n , m , i , x , y;
	unsigned z;
	n = read() , m = read();
	for(i = 1 ; i <= n ; i ++ ) si[i] = w[i] = sum[i] = mul[i] = 1;
	for(i = 1 ; i < n ; i ++ ) x = read() , y = read() , link(x , y);
	while(m -- )
	{
		scanf("%s" , str) , x = read() , y = read();
		switch(str[0])
		{
			case '+': z = (unsigned)read() , split(x , y) , cal(x , z , 1 , 0); break;
			case '-': cut(x , y) , x = read() , y = read() , link(x , y); break;
			case '*': z = (unsigned)read() , split(x , y) , cal(x , 0 , z , 0); break;
			default: split(x , y) , printf("%u\n" , sum[x]);
		}
	}
	return 0;
}

 

posted @ 2017-05-02 09:41  GXZlegend  阅读(259)  评论(0编辑  收藏  举报