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  • 【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
    " , sum[x]);
    		}
    	}
    	return 0;
    }

     

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  • 原文地址:https://www.cnblogs.com/GXZlegend/p/6794700.html
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