洛谷 P2073 送花

Description

洛谷 P2073 送花

Solution

无旋\(treap\) (\(fhq-treap\))

这道题有点小变化。

对于 \(insert\) 操作，就按花的价格分裂。判断是否有当前要插入的花的价格，如果有就直接合并回去，如果没有，就把当前花插入进去。

对于删除操作，我们就删除点 1，或点 tot。（显然一个最大，一个最小，删哪个根据题意）

这时我们要再写一个分裂函数，按子树大小分裂，然后不用合并了，相当于删除了，具体看代码吧。

注意：按子树大小分裂时，子树的根有点变化。

Code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#define ls(x) t[x].ch[0]
#define rs(x) t[x].ch[1]
#define ll long long

using namespace std;

const ll N = 1e5 + 10;
struct Tree{
	ll ch[2], siz, val, wei, cost;
}t[N];
ll n, m, tot, root;
ll a, b, c;
ll ans1, ans2;

inline void pushup(ll x){
	t[x].siz = t[ls(x)].siz + t[rs(x)].siz + 1;
}

inline void split(ll x, ll k, ll &a, ll &b){
	if(!x){
		a = b = 0;
		return;
	}
	if(t[ls(x)].siz >= k){
		b = x;
		split(ls(x), k, a, ls(x));
		pushup(x);
	}else{
		a = x;
		split(rs(x), k - t[ls(x)].siz - 1, rs(x), b);
		pushup(x);
	}
}

inline void split2(ll x, ll k, ll &a, ll &b){
	if(!x){
		a = b = 0;
		return;
	}
	if(t[x].cost <= k){
		a = x;
		split2(rs(x), k, rs(x), b);
	}else{
		b = x;
		split2(ls(x), k, a, ls(x));
	}
	pushup(x);
}

inline ll newnode(ll val, ll cost){
	t[++tot].val = val, t[tot].cost = cost;
	t[tot].siz = 1, t[tot].wei = rand();
	return tot;
}

inline ll merge(ll x, ll y){
	if(!x || !y) return x + y;
	if(t[x].wei < t[y].wei){
		rs(x) = merge(rs(x), y);
		pushup(x);
		return x;
	}else{
		ls(y) = merge(x, ls(y));
		pushup(y);
		return y;
	}
}

inline void insert(ll w, ll cost){
	split2(root, cost, a, b);
	split2(a, cost - 1, a, c);
	if(t[c].siz) root = merge(a, merge(c, b));
	else root = merge(a, merge(newnode(w, cost), b));
}

void dfs(ll x){
	if(!x) return;
	ans1 += t[x].val;
	ans2 += t[x].cost;
	dfs(ls(x)), dfs(rs(x));
}

signed main(){
	ll op, w, cost;
	while(scanf("%lld", &op)){
		if(op == -1) break;
		if(op == 1) {
			scanf("%lld%lld", &w, &cost);
			insert(w, cost);
		}
		if(op == 2) split(root, t[root].siz - 1, root, a);
		if(op == 3) split(root, 1, a, root);
	}
	dfs(root);
	printf("%lld %lld\n", ans1, ans2);
	return 0;	
}

End