鬼子进村 fhq-treap

鬼子进村 fhq-treap

题面

观察题目发现可用平衡树做：每次鬼子拆家即从平衡树中加入被拆的节点；每次村民修房子都向平衡树中删除该节点；每次查询时，只需要求出其后驱与前驱，易知nxt-pre-1为答案。

使用\(\text{fhq-treap}\)实现平衡树部分

#include <cstdio>
#include <cstdlib>
#define MAXN 50005
using namespace std;
struct nod{
	int sl,sr,val,rnd,sz;
} tre[MAXN];
void update(int x){
	tre[x].sz=1+tre[tre[x].sl].sz+tre[tre[x].sr].sz;
}
int merge(int x, int y){
	if(x==0||y==0) return x|y;
	if(tre[x].rnd<tre[y].rnd){
		tre[x].sr=merge(tre[x].sr, y);
		update(x);
		return x;
	}else{
		tre[y].sl=merge(x, tre[y].sl);
		update(y);
		return y;
	}
}
void split(int cur, int k, int &x, int &y){
	if(cur==0){x=y=0; return;}
	if(tre[cur].val<=k){
		x=cur;
		split(tre[cur].sr, k, tre[cur].sr, y);
	}else{
		y=cur;
		split(tre[cur].sl, k, x, tre[cur].sl);
	}
	update(cur);
}
int tot,rot,x,y,z;
int new_nod(int val){
	tre[++tot].val=val;
	tre[tot].sz=1;
	tre[tot].rnd=rand();
	return tot;
}
void add(int val){
	split(rot, val, x, y);
	rot=merge(merge(x, new_nod(val)), y);
}
void del(int val){
	split(rot, val, x, z);
	split(x, val-1, x, y);
	y=merge(tre[y].sl, tre[y].sr);
	rot=merge(merge(x, y), z);
}
int get_kth(int cur, int k){
	while(1){
		if(k<=tre[tre[cur].sl].sz) cur=tre[cur].sl;
		else if(k==tre[tre[cur].sl].sz+1) return cur;
		else k-=tre[tre[cur].sl].sz+1, cur=tre[cur].sr;
	}
}
int get_pre(int val){
	split(rot, val-1, x, y);
	int res=get_kth(x, tre[x].sz);
	rot=merge(x,y);
	return res;
}
int get_nxt(int val){
	split(rot, val, x, y);
	int res=get_kth(y, 1);
	rot=merge(x,y);
	return res;
}
int n,m;
int s[MAXN],top;
bool des[MAXN];
int main(){
	srand((unsigned)19270817);
	scanf("%d %d", &n, &m);
	add(0),add(n+1);
	while(m--){
		char opt;int t;
		scanf("\n%c ", &opt);
		if(opt=='D'){
			scanf("%d", &t);
			add(t);
			s[++top]=t;
			des[t]=1;
		}else if(opt=='Q'){
			scanf("%d", &t);
			if(des[t]) printf("0\n");
			else printf("%d\n", tre[get_nxt(t)].val-tre[get_pre(t)].val-1);
		}else if(opt=='R'){
			while(des[s[top]]==0) ++top;
			des[s[top]]=0;
			del(s[top--]);
		}else puts("Erro!");
	}
	return 0;
}