[CF762E]Radio stations

Posted 2020-11-07 jefflyy

题意：给出$n$个广播站，每个广播站有坐标$x_i$，广播半径$r_i$，广播频率$f_i$，求满足$i\lt j,\min(r_i,r_j)\geq|x_i-x_j|,|f_i-f_j|\leq k$的$(i,j)$对数

因为$f_i$很小，$k$更小，所以不妨考虑枚举$f$，那么频率的限制可以转化只考虑满足$f_j\in[f-k,f+k]$的$j$

现在考虑统计$x_i,r_i$的约束，将$f_i\in[f-k,f-1]\cup[f+1,f+k]$的$i$按$x_i-r_i$排序，再对$f_i=f$的$i$按$x_i$从小到大查询有多少满足约束的，用树状数组和以$x_i+r_i$为关键字的小根堆动态维护当前能广播到$i$的站（按排序顺序插入并删除小根堆中不满足约束的），直接在树状数组上查询$[x_i-r_i,x_i+r_i]$即可，频率相同的最后再统计即可

#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
using namespace std;
typedef long long ll;
struct sta{
	int x,r;
	sta(int a=0,int b=0){x=a;r=b;}
};
bool operator<(sta a,sta b){return a.x+a.r>b.x+b.r;}
bool cmpx(sta a,sta b){return a.x<b.x;}
bool cmpl(sta a,sta b){return a.x-a.r<b.x-b.r;}
vector<sta>f[10010],now,nei;
vector<sta>::iterator v1,v2;
map<int,int>pos;
map<int,int>::iterator it;
priority_queue<sta>q;
int k,N,maxf,T[100010];
int lowbit(int x){return x&-x;}
int query(int x){
	int s=0;
	while(x){
		s+=T[x];
		x-=lowbit(x);
	}
	return s;
}
int query(int l,int r){
	it=pos.upper_bound(r);
	if(it==pos.begin())return 0;
	it--;
	r=query(it->second);
	l=pos.lower_bound(l)->second;
	if(l>1)r-=query(l-1);
	return r;
}
void modify(int x,int v){
	while(x<=N){
		T[x]+=v;
		x+=lowbit(x);
	}
}
ll calc(int x){
	int i;
	ll ans;
	now=f[x];
	nei.clear();
	for(i=max(x-k,1);i<=min(x+k,maxf);i++){
		if(i!=x){
			for(v1=f[i].begin();v1!=f[i].end();v1++)nei.push_back(*v1);
		}
	}
	sort(now.begin(),now.end(),cmpx);
	sort(nei.begin(),nei.end(),cmpl);
	v2=nei.begin();
	ans=0;
	for(v1=now.begin();v1!=now.end();v1++){
		while(v2!=nei.end()&&v2->x-v2->r<=v1->x){
			q.push(*v2);
			modify(pos[v2->x],1);
			v2++;
		}
		while(!q.empty()&&q.top().x+q.top().rx){
			modify(pos[q.top().x],-1);
			q.pop();
		}
		ans+=query(v1->x-v1->r,v1->x+v1->r);
	}
	while(!q.empty()){
		modify(pos[q.top().x],-1);
		q.pop();
	}
	for(v1=now.begin();v1!=now.end();v1++){
		while(!q.empty()&&q.top().x+q.top().rx){
			modify(pos[q.top().x],-2);
			q.pop();
		}
		ans+=query(v1->x-v1->r,v1->x+v1->r);
		modify(pos[v1->x],2);
		q.push(*v1);
	}
	while(!q.empty()){
		modify(pos[q.top().x],-2);
		q.pop();
	}
	return ans;
}
int main(){
	int n,i,a,b,c;
	ll ans;
	scanf("%d%d",&n,&k);
	for(i=1;i<=n;i++){
		scanf("%d%d%d",&a,&b,&c);
		pos[a]=1;
		f[c].push_back(sta(a,b));
		maxf=max(maxf,c);
	}
	for(N=1,it=pos.begin();it!=pos.end();it++,N++)it->second=N;
	N--;
	ans=0;
	for(i=1;i<=maxf;i++)ans+=calc(i);
	printf("%I64d",ans>>1);
}