[洛谷][P1503][鬼子进村][Treap]

原题地址

题意就不解释了吧。

看到这道题，一开始YY把所有点加进treap里面，然后完全跟着操作走，应该是可做的。[在treap里面找连续的一段应该很简单，只要有一定代码能力都行|||||||吧。。。。]

可不幸的因为本人太弱，不想写，就不得不YY其它的做法，我们可以发现，被困士兵能通过的房子组成的序列的两个端点为被毁灭的两个房屋。[左端点有可能为零，即最左端，右端点可能为n+1，即最右端] 所以我们每次把被删除的端点加入treap，对于查询，我们查当前士兵的前驱和后继即可

上代码 :

#include
#include
#include
#include
#include
#include
#include

const int N = 5e4 + 7,INF = 0x7fffffff;

#define R121() ((rand()<<12)+(rand()<<7)+rand())
#include
#define R233() (R121() % 456454541 + R121() % 452112101 + R121() % 121231 + 1)
using std :: bitset;

int n,m,v;

class Treap {
private :
	struct Node {
		Node *son[2];
		int size,data,hr;
		Node () {}
		Node (int data,Node *fl) : data(data) { son[0] = son[1] = fl; size = 1; hr = R233(); }
		void update () { size = son[0] -> size + son[1] -> size + 1;}
	}*Null,*root,*pool,meme[N];
	
	int ball;
	
	void rotate (Node *&T,bool v) {
		Node *Tt = T -> son[v];
		T -> son[v] = Tt -> son[v^1];
		Tt -> son[v^1] = T;
		T -> update(); Tt -> update();
		T = Tt;
	}
	
	void Insert (Node *&T) {
		if(T == Null) { T = new (pool++) Node (ball,Null); return ; }
		bool v = T -> data < ball;
		Insert (T -> son[v]);
		if(T -> son[v] -> hr < T -> hr) rotate (T , v);
		else T -> update();
	}
	
	void Delete (Node *&T) {
		if(T == Null) return;
		if(T -> data == ball) {
			if(T -> son[0] == Null || T -> son[1] == Null) {
				bool v = T -> son[1] != Null;
				T = T -> son[v];
				return ;
			}
			bool v = T -> son[0] -> hr > T -> son[1] -> hr;
			rotate (T , v);
			Delete (T -> son[v^1]);
		} else {
			bool v = ball > T -> data;
			Delete (T -> son[v]);
		}
		T -> update();
	}
	
	int Subsequent (Node *&T) {
		if(T == Null) return n+1;
		if(T -> data <= ball) return Subsequent (T -> son[1]);
//		if(T -> data + 1 <= ball) return T -> data;
//		if(T -> data < ball) 
		int k = Subsequent (T -> son[0]);
		return k == n + 1 ? T -> data : k;
	}
	int Precursor (Node *&T) {
		if(T == Null) return 0;
		if(T -> data >= ball) return Precursor (T -> son[0]);
		int k = Precursor (T -> son[1]);
		return k == 0 ? T -> data : k;
	}
	
	
public :
	Treap () { Null = new Node(); Null -> size = 0; }
	void clear () { pool = meme; root = Null; }
	void Ins (int xxx) { ball = xxx; Insert(root); }
	void Del (int xxx) { ball = xxx; Delete(root); }
	int Sub (int xxx) { ball = xxx; return Subsequent(root); }
	int Pre (int xxx) { ball = xxx; return Precursor(root); }
}treap;

char opt[N];
bitsetvis;
int stk[N],top;

int main () {
	scanf("%d%d",&n,&m); treap.clear();
	for(int i=1;i<=m;++i) {
		scanf("%s",opt);
		if(opt[0] == 'R') treap.Del(stk[top--]),vis.flip(stk[top+1]);
		else {
			scanf("%d",&v);
			if(opt[0] == 'D') { if(!vis.test(v)) treap.Ins(v),vis.flip(v),stk[++top]=v; }
			else if(opt[0] == 'Q') {
				if(vis.test(v)) puts("0");
				else printf("%d\n",	treap.Sub(v) - treap.Pre(v) - 1);
			}
		}
	}
	return 0;	
}

That is all. Thank you for watching!

转载于:https://www.cnblogs.com/dcoi-king/p/5353660.html