gym 101480 Problem C: Cow Confinement 题解

gym 101480 Problem C: Cow Confinement 题解

算法知识：dp+扫描线。

从下往上扫描，维护一个线段树就可以了。线段树支持区间清零，区间加和单点查询。

Warning：扫描线预处理排序的部分需要注意先后顺序。

/*
{
######################
#       Author       #
#        Gary        #
#        2020        #
######################
*/
#include<bits/stdc++.h>
#define rb(a,b,c) for(int a=b;a<=c;++a)
#define rl(a,b,c) for(int a=b;a>=c;--a)
#define LL long long
#define IT iterator
#define PB push_back
#define II(a,b) make_pair(a,b)
#define FIR first
#define SEC second
#define FREO freopen("check.out","w",stdout)
#define rep(a,b) for(int a=0;a<b;++a)
#define SRAND mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define random(a) rng()%a
#define ALL(a) a.begin(),a.end()
#define POB pop_back
#define ff fflush(stdout)
#define fastio ios::sync_with_stdio(false)
#define check_min(a,b) a=min(a,b)
#define check_max(a,b) a=max(a,b)
using namespace std;
//inline int read(){
//    int x=0;
//    char ch=getchar();
//    while(ch<'0'||ch>'9'){
//        ch=getchar();
//    }
//    while(ch>='0'&&ch<='9'){
//        x=(x<<1)+(x<<3)+(ch^48);
//        ch=getchar();
//    }
//    return x;
//}
const int INF=0x3f3f3f3f;
typedef pair<int,int> mp;
/*}
*/
const int N=1<<20;
bool asi[N+N];//是否都设置成0
int delta[N+N];//增量
void pd(int now){
	if(asi[now]){
		asi[now<<1]=asi[now<<1|1]=1;
		delta[now<<1]=delta[now<<1|1]=delta[now];
	}
	else{
		delta[now<<1]+=delta[now];
		delta[now<<1|1]+=delta[now];
	}
	asi[now]=0;
	delta[now]=0;
}
void Assign(int a,int b,int now=1,int l=1,int r=N+1){
	if(r<=a||l>=b){
		return;
	}
	if(r<=b&&l>=a){
		delta[now]=0;
		asi[now]=1;
		return ;
	}
	pd(now);
	int mid=(l+r)>>1;
	Assign(a,b,now<<1,l,mid);
	Assign(a,b,now<<1|1,mid,r);
}
void Add(int a,int b,int val,int now=1,int l=1,int r=N+1){
	if(r<=a||l>=b) return ;
	if(r<=b&&l>=a){
		delta[now]+=val;
		return;
	}
	pd(now);
	int mid=(l+r)>>1;
	Add(a,b,val,now<<1,l,mid);
	Add(a,b,val,now<<1|1,mid,r);
}
int query(int pos){
	pos+=N-1;
	int rest=0;
	while(pos){
		if(asi[pos]){
			rest=0;
		}	
		rest+=delta[pos];
		pos>>=1;
	}
	return rest;
}
int answer[N],tmpanswer[N]/*记录每一个fence右下角的答案*/;
int f,m,n;
set<int> wall;
struct Event{
	int ty;//种类
	/*
	1. 上边界 
	2. 下边界
	3. 花 
	4. 牛 
	*/ 
	int r,c,len,id;
	Event(){}
	Event(int T,int X,int Y){
		ty=T;
		r=X;
		c=Y;
	} 
	Event(int T,int X,int Y,int L,int I){
		ty=T;
		r=X;
		c=Y;
		len=L;
		id=I;
	} 
	bool operator < (Event oth){
		if(r!=oth.r) return r>oth.r;
		if((!(ty<=2&&oth.ty<=2))&&ty!=oth.ty) return ty<oth.ty;
		return c<oth.c;
	}
};
vector<Event> v;
int main(){
	scanf("%d",&f);
	rb(i,1,f){
		int r1,c1,r2,c2;
		scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
		r1--;
		v.PB(Event(1,r1,c1,c2-c1+1,i));
		v.PB(Event(2,r2,c1,c2-c1+1,i));
	}
	scanf("%d",&m);
	rb(i,1,m){
		int r,c;
		scanf("%d%d",&r,&c);
		v.PB(Event(3,r,c));
	}
	scanf("%d",&n); 
	rb(i,1,n){
		int r,c;
		scanf("%d%d",&r,&c);
		v.PB(Event(4,r,c,0,i));
	}
	sort(ALL(v));
	for(int i=0;i<v.size();++i){
		int j;
		for(j=i;j<v.size()&&v[j].r==v[i].r;++j){
			Event Eve=v[j];
			if(Eve.ty==1){
				Assign(Eve.c,Eve.c+Eve.len);
				wall.erase(Eve.c-1);
				wall.erase(Eve.c+Eve.len-1);
				int tmp=query(Eve.c+Eve.len);
				tmp-=tmpanswer[Eve.id];
				auto can=wall.lower_bound(Eve.c);
				if(can==wall.begin()) Add(1,Eve.c+Eve.len,tmp);
				else{
					--can;
					Add((*can)+1,Eve.c+Eve.len,tmp);
				}
				Add(Eve.c,Eve.c+Eve.len,tmpanswer[Eve.id]);
			}
			if(Eve.ty==2){
				wall.insert(Eve.c-1);
				wall.insert(Eve.c+Eve.len-1);
				Assign(Eve.c,Eve.c+Eve.len);
				tmpanswer[Eve.id]=query(Eve.c+Eve.len);
			}
			if(Eve.ty==3){
				auto can=wall.lower_bound(Eve.c);
				if(can==wall.begin()) Add(1,Eve.c+1,1);
				else{
					--can;
					Add((*can)+1,Eve.c+1,1);
				}
			}
			if(Eve.ty==4){
				answer[Eve.id]=query(Eve.c);
			}
		}
		i=j-1;
	}
	rb(i,1,n) printf("%d
",answer[i]);
	return 0;
}/*

4
2 2 8 4
1 9 4 10
6 7 9 9
3 3 7 3

1
2 2 8 4
9
3 4
8 4
11 5
10 7
10 8
9 8
2 8
4 11
9 11
1
1 1

*/