[BZOJ 1036] [ZJOI2008] 树的统计Count 【Link Cut Tree】

题目链接:BZOJ – 1036

 

题目分析

这道题可以用树链剖分,块状树等多种方法解决,也可以使用 LCT。

修改某个点的值时,先将它 Splay 到它所在的 Splay 的根,然后修改它的值,再将它 Update 一下。

(1)

询问 x, y 两点之间的路径时,假设 x 是深度小的那一个,先 Access(x) ,然后再 Access(y) 的返回值就是 x, y 的 LCA 。

这时从 x 到 LCA 的路径已经在 LCA 处断开了。我们将 x Splay 一下,然后就是 x 所在的 Splay, LCA, Son[LCA][1] 这 3 部分组成了 x, y 的路径。

要特判 x == LCA 的情况。

(2)

在询问 x, y 两点之间的路径时,也可以用下面的方法:

将 x 变为这棵树的根,然后 Access(y) ,就是求 y 到根的路径了。

为了实现把一个点变为根,需要将这个点到当前树根的路径翻转,就是说,Access(x) , Splay(x), Reverse(x) 。

然后需要注意的就是翻转标记的下传,当要改变 Splay 的形态之前,就需要将涉及到的点 PushDown 一下,使他们没有标记。

比如 Rotate 的时候,或者 Access 中断掉 x 的右子树之前。一定要注意!

Rotate(x) 时, y = Father[x] ,要先 PushDown(y); PushDown(x); 然后再做下面的操作。

 

代码

不换根的代码:

#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 = 30000 + 5, INF = 999999999;

int n, q, Ans;
int A[MaxN], Father[MaxN], Sum[MaxN], Max[MaxN], Son[MaxN][2], Depth[MaxN];

bool isRoot[MaxN];

struct Edge
{
	int v;
	Edge *Next;
} E[MaxN * 2], *P = E, *Point[MaxN];

inline void AddEdge(int x, int y)
{
	++P; P -> v = y;
	P -> Next = Point[x]; Point[x] = P;
}

void DFS(int x, int Fa, int Dep)
{
	Father[x] = Fa; Depth[x] = Dep;
	for (Edge *j = Point[x]; j; j = j -> Next)
	{
		if (j -> v == Fa) continue;
		DFS(j -> v, x, Dep + 1);
	}
}

inline int gmin(int a, int b) {return a < b ? a : b;}
inline int gmax(int a, int b) {return a > b ? a : b;}

inline void Update(int x)
{
	Max[x] = gmax(A[x], gmax(Max[Son[x][0]], Max[Son[x][1]]));
	Sum[x] = A[x] + Sum[Son[x][0]] + Sum[Son[x][1]];
}

void Rotate(int x, int f)
{
	int y = Father[x];
	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])
		{
			if (x == Son[y][0]) Rotate(x, 1);
			else Rotate(x, 0);
			break;
		}
		if (y == Son[Father[y]][0])
		{
			if (x == Son[y][0]) 
			{
				Rotate(y, 1);
				Rotate(x, 1);
			}
			else
			{
				Rotate(x, 0);
				Rotate(x, 1);
			}
		}
		else
		{
			if (x == Son[y][1])
			{
				Rotate(y, 0);
				Rotate(x, 0);
			}
			else
			{
				Rotate(x, 1);
				Rotate(x, 0);
			}
		}
	}
}

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

char Str[10];

int main()
{
	scanf("%d", &n);
	int a, b, LCA;
	for (int i = 1; i <= n - 1; ++i)
	{
		Read(a); Read(b);
		AddEdge(a, b); AddEdge(b, a);
	}
	DFS(1, 0, 0);
	for (int i = 1; i <= n; ++i)
	{
		Read(A[i]);
		isRoot[i] = true;
		Sum[i] = Max[i]= A[i];
	}
	Sum[0] = 0; Max[0] = -INF;
	scanf("%d", &q);
	for (int i = 1; i <= q; ++i)
	{
		scanf("%s", Str);
		Read(a); Read(b);
		if (strcmp(Str, "CHANGE") == 0)
		{
			Splay(a); 
			A[a] = b; 
			Update(a);
		}
		else if (strcmp(Str, "QMAX") == 0)
		{
			if (Depth[a] > Depth[b]) swap(a, b);
			Access(a);
			LCA = Access(b);
			Splay(a);
			if (a == LCA) Ans = gmax(A[LCA], Max[Son[LCA][1]]);
			else Ans = gmax(A[LCA], gmax(Max[a], Max[Son[LCA][1]]));
			printf("%d\n", Ans);
		}
		else if (strcmp(Str, "QSUM") == 0)
		{
			if (Depth[a] > Depth[b]) swap(a, b);
			Access(a);
			LCA = Access(b);
			Splay(a);
			if (a == LCA) Ans = A[LCA] + Sum[Son[LCA][1]];
			else Ans = A[LCA] + Sum[a] + Sum[Son[LCA][1]];
			printf("%d\n", Ans);
		}
	}
	return 0;
}

换根的代码:

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#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;
}

inline int gmin(int a, int b) {return a < b ? a : b;}
inline int gmax(int a, int b) {return a > b ? a : b;}

const int MaxN = 30000 + 5, INF = 999999999;

int n, q;
int A[MaxN], Father[MaxN], Son[MaxN][2], Max[MaxN], Sum[MaxN];

bool isRoot[MaxN], Rev[MaxN];

struct Edge
{
	int v;
	Edge *Next;
} E[MaxN * 2], *P = E, *Point[MaxN];

inline void AddEdge(int x, int y)
{
	++P; P -> v = y;
	P -> Next = Point[x]; Point[x] = P;
}

void DFS(int x, int Fa)
{
	Father[x] = Fa;
	for (Edge *j = Point[x]; j; j = j -> Next)
	{
		if (j -> v == Fa) continue;
		DFS(j -> v, x);
	}
}

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

inline void PushDown(int x)
{
	if (!Rev[x]) return;
	Rev[x] = false;
	if (Son[x][0]) Reverse(Son[x][0]);	
	if (Son[x][1]) Reverse(Son[x][1]);	
}

inline void Update(int x)
{
	Sum[x] = Sum[Son[x][0]] + Sum[Son[x][1]] + A[x];
	Max[x] = gmax(A[x], gmax(Max[Son[x][0]], Max[Son[x][1]]));
}

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][1 ^ f] = 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);
		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", &n);
	int a, b, c;
	for (int i = 1; i <= n - 1; ++i)
	{
		Read(a); Read(b);
		AddEdge(a, b); AddEdge(b, a);
	}
	for (int i = 1; i <= n; ++i) Read(A[i]);
	DFS(1, 0);
	Max[0] = -INF; Sum[0] = 0;
	for (int i = 1; i <= n; ++i) 
	{
		isRoot[i] = true;
		Sum[i] = Max[i] = A[i];
	}
	scanf("%d", &q);
	char Str[10];
	int Ans, LCA;
	for (int i = 1; i <= q; ++i)
	{
		scanf("%s", Str);
		Read(a); Read(b);
		if (strcmp(Str, "CHANGE") == 0)
		{	
			Splay(a); 
			A[a] = b;
			Update(a);
 		}
		else 
		{
			Make_Root(a);
			c = Access(b);
			if (strcmp(Str, "QMAX") == 0) printf("%d\n", Max[c]);
			else printf("%d\n", Sum[c]);
		}
	}
	return 0;
}

  

    原文作者:Online Judge
    原文地址: https://www.cnblogs.com/JoeFan/p/4433019.html
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