原题链接
考察:线段树
思路:
  和校门外的树(增强版)那道题差不多,我们不用管哪些瓶子有花,哪些没有.只要add,范围内从小到大 = 1.只要删除,sum[l,r] = 0.

Code

#include <iostream> 
#include <cstring>
using namespace std;
const int N = 50011,INF = 0x3f3f3f3f;
struct Node{
	int l,r,sum,tag;
}tr[N<<2];
int n,m,ans_l,ans_r,f;
void build(int u,int l,int r)
{
	tr[u] = {l,r,0,0};
	if(l==r) return;
	int mid = l+r>>1;
	build(u<<1,l,mid),build(u<<1|1,mid+1,r);
}
void push_up(int u)
{
	tr[u].sum = tr[u<<1].sum+tr[u<<1|1].sum;
}
int get(int u)
{
	return tr[u].r-tr[u].l+1;
}
void push_down(int u)
{
	if(tr[u].tag==1)
	{
		tr[u<<1].sum = get(u<<1);
		tr[u<<1|1].sum = get(u<<1|1);
		tr[u<<1].tag = tr[u<<1|1].tag = tr[u].tag;
		tr[u].tag = 0;
	}
	if(tr[u].tag==-1)
	{
		tr[u<<1].sum = 0,tr[u<<1|1].sum = 0;
		tr[u<<1].tag = tr[u<<1|1].tag = tr[u].tag;
		tr[u].tag = 0;
	}
}
int del(int u,int l,int r)
{
	int res = 0;
	if(tr[u].l>=l&&tr[u].r<=r)
	{
		res+=tr[u].sum;
		tr[u].sum = 0;
		tr[u].tag = -1;
		return res;
	}
	push_down(u);
	int mid = tr[u].l+tr[u].r>>1;
	if(l<=mid) res+=del(u<<1,l,r);
	if(mid<r) res+=del(u<<1|1,l,r);
	push_up(u);
	return res;
}
void add(int u,int l,int r)
{
	if(!f) return;
	if(tr[u].sum==get(u)) return;
	if(tr[u].l==tr[u].r)
	{
		if(!tr[u].sum) ans_l = min(ans_l,tr[u].l),ans_r = max(ans_r,tr[u].r),f-=1;
		tr[u].sum = 1;
		return;
	}
	if(tr[u].l>=l&&tr[u].r<=r&&f>=get(u)&&!tr[u].sum) 
	{
		tr[u].sum = get(u);
		f-=get(u);
		ans_l = min(ans_l,tr[u].l);
		ans_r = max(ans_r,tr[u].r); 
		tr[u].tag = 1;
		return;
	}
	push_down(u);
	int mid = tr[u].l+tr[u].r>>1;
	if(l<=mid) add(u<<1,l,r);
	if(mid<r) add(u<<1|1,l,r);
	push_up(u);
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		build(1,1,n);
		while(m--)
		{
			int op,l,r;
			scanf("%d%d%d",&op,&l,&r);
			l++,r++;
			if(op==2) { printf("%d\n",del(1,l,r));continue; }
			ans_l = INF,ans_r = -INF;
			f = --r;
			add(1,l,n);
			if(f==r) puts("Can not put any one.") ;
			else printf("%d %d\n",ans_l-1,ans_r-1);
		}
		printf("\n");
	}
	return 0;
}

标签：HDU,get,int,sum,tr,&&,ans,Flowers,4614
来源： https://www.cnblogs.com/newblg/p/14799569.html