NOIP 模拟 序列操作

题意：

一开始有n个非负整数h[i]，接下来会进行m次操作，第i次会给出一个数c[i]，要求选出c[i]个大于0的数并将它们-1，问最多可以进行多少次？

分析：

首先一个显然的贪心就是每次都将最大的c[i]个数-1，于是就可以用无旋式treap来维护，基本操作中split_k和split_v都使用普通的merge，但在提取区间并打完标记后，因为整个序列的单调性发生改变，需要使用启发式合并（只在修改过后使用）。

code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<ctime>
#include<queue>
using namespace std;
template<typename T>
inline void read(T &x) {
    T i = 0, f = 1;
    char ch = getchar();
    for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
    if(ch == '-') f = -1, ch = getchar();
    for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
    x = i * f;
}
template<typename T>
inline void wr(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x > 9) wr(x / 10);
    putchar(x % 10 + '0');
}
const int N = 1e6 + 50, OO = 0x3f3f3f3f;
int n, m, h[N], c[N];
inline int Rand(){
	static int RAND_VAL = 1388593021;
	return RAND_VAL += RAND_VAL << 2 | 1;
}
#define SZ(x) (x?x->sze:0)
#define V(x) (x?x->val:OO)
struct node{
	node *lc, *rc;
	int sze, pri, val, tag;
	node():lc(NULL), rc(NULL){}
	inline void add(int v){
		val += v, tag += v;
	}
	inline void pushDown(){
		if(tag != 0){
			if(lc) lc->add(tag);
			if(rc) rc->add(tag);
			tag = 0;
		}
	}
	inline node* upt(){
		sze = SZ(lc) + SZ(rc) + 1;
		return this;
	}
	inline void print(){
		pushDown();
		if(lc) lc->print();
		cout<< val<<" ";
		if(rc) rc->print();
	}
}pool[N], *tail = pool, *rt = NULL;
inline node* newNode(int v){
	node *x = tail++;
	x->val = v;
	x->lc = x->rc = NULL;
	x->pri = Rand();
	x->tag = 0;
	x->sze = 1;
	return x;
}
inline node* Merge_o(node *x, node *y){
	if(!x) return y;
	if(!y) return x;
	node *L, *R;
	if(x->pri < y->pri){
		x->pushDown();
		x->rc = Merge_o(x->rc, y);
		return x->upt();
	}
	else {
		y->pushDown();
		y->lc = Merge_o(x, y->lc);
		return y->upt();
	}
}
inline void Split_v(node *x, int v, node *&L, node *&R);
inline node* Merge(node *x, node *y){
	if(!x) return y;
	if(!y) return x;
	x->pushDown(), y->pushDown();
	if(x->pri > y->pri) swap(x, y);
	node *L, *R;
	Split_v(y, x->val, L, R);
	x->lc = Merge(x->lc, L);
	x->rc = Merge(x->rc, R);
	x->upt();
	return x;
}
inline void Split_k(node *x, int k, node *&L, node *&R){
	if(x == NULL){L = NULL, R = NULL; return;}
	x->pushDown();
	if(SZ(x->lc) < k){
		Split_k(x->rc, k - SZ(x->lc) - 1, L, R);
		x->rc = NULL; x->upt();
		L = Merge_o(x, L);
	}
	else{
		Split_k(x->lc, k, L, R);
		x->lc = NULL; x->upt();
		R = Merge_o(R, x);
	}
}
inline void Split_v(node *x, int v, node *&L, node *&R){
	if(x == NULL){L = NULL, R = NULL; return;}
	x->pushDown();
	if(V(x) <= v){
		Split_v(x->rc, v, L, R);
		x->rc = NULL; x->upt();
		L = Merge_o(x, L);
	}
	else{
		Split_v(x->lc, v, L, R);
		x->lc = NULL, x->upt();
		R = Merge_o(R, x);
	}
}
inline node* build(){
	static node* stk[N];
	node* pre;
	int top = 0;
	for(register int i = 1; i <= n; i++){
		node *x = newNode(h[i]);
		pre = NULL;
		while(top && stk[top]->pri > x->pri)
			pre = stk[top]->upt(),  stk[top--] = NULL;
		if(stk[top]) stk[top]->rc = x;
		x->lc = pre; stk[++top] = x;
	}
	while(top) stk[top--]->upt();
	return stk[1];
}
int main(){
	freopen("h.in", "r", stdin);
	read(n), read(m);
	for(register int i = 1; i <= n; i++) read(h[i]);
	for(register int i = 1; i <= m; i++) read(c[i]);
	sort(h + 1, h + n + 1);
	rt = build();
	for(register int i = 1; i <= m; i++){
		if(!c[i]) continue;
		if(c[i] > n){
			wr(i - 1);
			return 0;
		}
		node *L, *R, *p, *q;
		Split_v(rt, 0, L, R);
		if(SZ(R) >= c[i]){
			Split_k(R, SZ(R) - c[i], p, q);
			if(q) q->add(-1);
			R = Merge(p, q);
		}
		else{
			wr(i - 1);
			return 0;
		}
		rt = Merge(L, R);
	}
	wr(m);
	return 0;
}