[BZOJ 3531] [Sdoi2014] 旅行 【离线+LCT】

题目链接:BZOJ - 3531

 

题目分析

题目询问一条路径上的信息时,每次询问有某种特定的文化的点。

每个点的文化就相当于一种颜色,每次询问一条路径上某种颜色的点的信息。

可以使用离线算法, 类似于“郁闷的小 J ” 那道题目。将各种操作和询问按照颜色为第一关键字,时间为第二关键字排序。

那么修改颜色的操作就相当于在原颜色中是删点,在新颜色中是加点。

处理完一种颜色的操作后,要将这个颜色的点都做一次删除操作,这样,对于处理下一种颜色,树就又是空的了。

这种题,思考的时候有点晕,写代码的时候非常愉悦。

 

代码

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

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 = 100000 + 5, MaxM = 100000 + 5, INF = 999999999;

int n, m, QTop, ATot;
int W[MaxN], C[MaxN], WT[MaxN], CT[MaxN], Ans[MaxM], Father[MaxN], T[MaxN], Sum[MaxN], Max[MaxN], Son[MaxN][2];

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;
}

struct ES
{
	int Type, Col, Pos, Num, x, y, TL, AIdx; 
	/*
		TypeList:
			1 : Change City-Pos to Num
			2 : Qsum of the path(x, y)
			3 : Qmax of the path(x, y)
	*/
} Q[MaxN * 2 + MaxM * 2];

inline bool Cmp(ES e1, ES e2)
{
	if (e1.Col != e2.Col) return e1.Col < e2.Col;
	return e1.TL < e2.TL;
}

queue<int> Qe;
 
void BFS()
{
	while (!Qe.empty()) Qe.pop();
	Qe.push(1); Father[1] = 0;
	int x;
	while (!Qe.empty()) 
	{
		x = Qe.front(); Qe.pop();
		for (Edge *j = Point[x]; j; j = j -> Next)
		{
			if (j -> v == Father[x]) continue;
			Father[j -> v] = x;
			Qe.push(j -> v);
		}
	}
}

/************************ LCT Start ****************************/

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

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 int GetDir(int x) {return x == Son[Father[x]][0] ? 0 : 1;}

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

inline 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;
}

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

/************************ LCT End ****************************/

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i)
	{
		Read(W[i]); Read(C[i]);
		WT[i] = W[i]; CT[i] = C[i];
	}
	int a, b;
	for (int i = 1; i <= n - 1; ++i)
	{
		Read(a); Read(b);
		AddEdge(a, b); AddEdge(b, a);
	}
	char Str[5];
	for (int i = 1; i <= n; ++i)
	{
		++QTop; Q[QTop].TL = 0; Q[QTop].Col = CT[i];
		Q[QTop].Type = 1; Q[QTop].Pos = i; Q[QTop].Num = WT[i];
	}
	for (int i = 1; i <= m; ++i)
	{
		scanf("%s", Str);
		Read(a); Read(b);
		if (strcmp(Str, "CC") == 0)
		{
			++QTop; Q[QTop].TL = i; Q[QTop].Col = CT[a];
			Q[QTop].Type = 1; Q[QTop].Pos = a; Q[QTop].Num = 0;
			++QTop; Q[QTop].TL = i; Q[QTop].Col = b;
			Q[QTop].Type = 1; Q[QTop].Pos = a; Q[QTop].Num = WT[a];
			CT[a] = b;		
		}
		else if (strcmp(Str, "CW") == 0)
		{
			++QTop; Q[QTop].TL = i; Q[QTop].Col = CT[a];
			Q[QTop].Type = 1; Q[QTop].Pos = a; Q[QTop].Num = b;
			WT[a] = b;
		}
		else if (strcmp(Str, "QS") == 0)
		{
			++QTop; Q[QTop].TL = i; Q[QTop].Col = CT[a];
			Q[QTop].Type = 2; Q[QTop].x = a; Q[QTop].y = b; Q[QTop].AIdx = ++ATot;
		}
		else if (strcmp(Str, "QM") == 0)
		{
			++QTop; Q[QTop].TL = i; Q[QTop].Col = CT[a];
			Q[QTop].Type = 3; Q[QTop].x = a; Q[QTop].y = b; Q[QTop].AIdx = ++ATot;
		}
	}
	for (int i = 1; i <= n; ++i)
	{
		++QTop; Q[QTop].TL = INF; Q[QTop].Col = CT[i];
		Q[QTop].Type = 1; Q[QTop].Pos = i; Q[QTop].Num = 0;
	}
	sort(Q + 1, Q + QTop + 1, Cmp);
	BFS();
	for (int i = 1; i <= n; ++i)	
	{
		isRoot[i] = true; Rev[i] = false;
		T[i] = 0; Max[i] = 0; Sum[i] = 0;
	}
	int t;
	for (int i = 1; i <= QTop; ++i)
	{
		if (Q[i].Type == 1)
		{
			Splay(Q[i].Pos);
			T[Q[i].Pos] = Q[i].Num;
			Update(Q[i].Pos);
		}
		else
		{
			Make_Root(Q[i].x);
			t = Access(Q[i].y);
			if (Q[i].Type == 2) Ans[Q[i].AIdx] = Sum[t];
			else Ans[Q[i].AIdx] = Max[t];	
		}
	}
	for (int i = 1; i <= ATot; ++i) printf("%d\n", Ans[i]);
	return 0;
}

  

 

posted @ 2015-04-25 21:27  JoeFan  阅读(268)  评论(0编辑  收藏  举报