[BZOJ - 2631] tree 【LCT】

题目链接:BZOJ – 2631

 

题目分析

LCT,像线段树区间乘,区间加那样打标记。

这道题我调了一下午。

提交之后TLE了,我一直以为是写错了导致了死循环。

于是一直在排查错误。直到..

直到我看了hzwer的博客,就一句话:“其实这题不需要开long long。。。只要unsigned int,不然可能会T”。

纳尼?!TLE是因为常数问题?于是我将 long long 改成了 unsighed int ,然后…AC。

我…………….!

 

long long 不能随便用啊…

 

代码

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

inline void Read(int &Num)
{
	char c = getchar();
	bool Neg = false;
	while (c < '0' || c > '9') 
	{
		if (c == '-') Neg = true;
		c = getchar();
	}
	Num = c - '0'; c = getchar();
	while (c >= '0' && c <= '9')
	{
		Num = Num * 10 + c - '0';
		c = getchar();
	}
	if (Neg) Num = -Num;
}

const int MaxN = 100000 + 5, Mod = 51061;

typedef unsigned int LL;

int n, q;
int Father[MaxN], Son[MaxN][2], Size[MaxN];

LL T[MaxN], Mul[MaxN], Plus[MaxN], A[MaxN]; 

bool Rev[MaxN], isRoot[MaxN];

inline void Update(int x)
{
	T[x] = (T[Son[x][0]] + T[Son[x][1]] + A[x]) % Mod;
	Size[x] = Size[Son[x][0]] + Size[Son[x][1]] + 1;
}

inline void Reverse(int x)
{
	Rev[x] = !Rev[x];
	swap(Son[x][0], Son[x][1]);
}

inline void Paint_Mul(int x, LL Num)
{
	Mul[x] = Mul[x] * Num % Mod;
	Plus[x] = Plus[x] * Num % Mod;
	T[x] = T[x] * Num % Mod;
	A[x] = A[x] * Num % Mod;
}

inline void Paint_Plus(int x, LL Num)
{
	Plus[x] = (Plus[x] + Num) % Mod;
	T[x] = (T[x] + Num * Size[x] % Mod) % Mod;
	A[x] = (A[x] + Num) % Mod;
}

inline void PushDown(int x)
{
	if (Rev[x])
	{
		Rev[x] = false;
		if (Son[x][0]) Reverse(Son[x][0]);
		if (Son[x][1]) Reverse(Son[x][1]);
 	}
 	if (Mul[x] != 1)
 	{
 		if (Son[x][0]) Paint_Mul(Son[x][0], Mul[x]);
 		if (Son[x][1]) Paint_Mul(Son[x][1], Mul[x]);
 		Mul[x] = 1;
 	}
 	if (Plus[x] != 0)
 	{
 		if (Son[x][0]) Paint_Plus(Son[x][0], Plus[x]);
 		if (Son[x][1]) Paint_Plus(Son[x][1], Plus[x]);
 		Plus[x] = 0;
 	}
}

void Rotate(int x)
{
	int y = Father[x], f;
	PushDown(y); PushDown(x);
	if (x == Son[y][0]) f = 1;
	else f = 0;
	if (isRoot[y])
	{
		isRoot[y] = false;
		isRoot[x] = true;
	}
	else
	{
		if (y == Son[Father[y]][0]) Son[Father[y]][0] = x;
		else Son[Father[y]][1] = x;
	}
	Father[x] = Father[y];
	Son[y][f ^ 1] = Son[x][f];
	if (Son[x][f]) Father[Son[x][f]] = y;
	Son[x][f] = y;
	Father[y] = x;
	Update(y);
	Update(x);
}

void Splay(int x)
{
	int y;
	while (!isRoot[x])
	{
		y = Father[x];
		if (isRoot[y])
		{
			Rotate(x);
			break;
		}
		if (y == Son[Father[y]][0])
		{
			if (x == Son[y][0])
			{
				Rotate(y);
				Rotate(x);
			}
			else
			{
				Rotate(x);
				Rotate(x);
			}
		}
		else
		{
			if (x == Son[y][1])
			{
				Rotate(y);
				Rotate(x);
			}
			else
			{
				Rotate(x);
				Rotate(x);
			}
		}
	}
}

int Access(int x)
{
	int y = 0;
	while (x != 0)
	{
		Splay(x);
		PushDown(x);
		if (Son[x][1]) isRoot[Son[x][1]] = true;
		Son[x][1] = y;
		if (y) isRoot[y] = false;
		Update(x);
		y = x;
		x = Father[x];
	}
	return y;
}

void Make_Root(int x)
{
	int t = Access(x);
	Reverse(t);
}

int main()
{
	scanf("%d%d", &n, &q);
	int a, b;	
	for (int i = 1; i <= n; ++i)
	{
		A[i] = T[i] = 1;
		Mul[i] = 1; Plus[i] = 0;
		isRoot[i] = true;
		Size[i] = 1;
	}
	for (int i = 1; i <= n - 1; ++i)
	{
		Read(a); Read(b);
		Make_Root(a);
		Splay(a);
		Father[a] = b;
	}
	char Str[5];
	int x, y, Num, t;
	for (int i = 1; i <= q; ++i)
	{
		scanf("%s", Str);
		switch (Str[0])
		{
			case '+' : 
				Read(x); Read(y); Read(Num);
				Make_Root(x);
				t = Access(y);
				Paint_Plus(t, (LL)Num);
				break;
			
			case '-' :
				Read(x); Read(y);
				Make_Root(x);
				Access(y);
				Splay(y);
				PushDown(y);
				isRoot[Son[y][0]] = true;
				Father[Son[y][0]] = 0;
				Son[y][0] = 0;
				Update(y);
				Read(x); Read(y);
				Make_Root(x);
				Splay(x);
				Father[x] = y;
				break;
				
			case '*' :
				Read(x); Read(y); Read(Num);
				Make_Root(x);
				t = Access(y);
				Paint_Mul(t, (LL)Num);
				break;
			
			case '/' : 
				Read(x); Read(y);
				Make_Root(x);
				t = Access(y);
				printf("%d\n", (int)(T[t] % Mod));	
				break;
		}
	}
	return 0;
}

  

    原文作者:Online Judge
    原文地址: https://www.cnblogs.com/JoeFan/p/4448199.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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