[haoi2014]遥感监测

Posted 2020-09-01 CHADLZX

蛤省前几年的省选题有点水啊.

对于每个点,可以用r和y算出能覆盖到它的望远镜的的坐标的左右范围,算出这个后,我们发现,问题转化成了,一条直线上有n个线段,如何用最少的点放在坐标轴上使所有线段上至少有一个点.

对于这个问题,首先观察到包含别的线段的线段不用考虑,先去掉.

去完后,发现整个轴成了线段不断交替的过程,然后贪心即可.

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<string>
#include<map>
#include<cstdlib>
#include<cmath>
#include<cstdio>
#include<ctime>
#include<queue>
using namespace std;
#define LL long long
#define FILE "dealing"
#define eps 1e-10
#define db double
#define pii pair<int,int>
#define up(i,j,n) for(int i=j;i<=n;i++)
int read(){
	int x=0,f=1,ch=getchar();
	while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch>=‘0‘&&ch<=‘9‘)x=(x<<1)+(x<<3)+ch-‘0‘,ch=getchar();
	return x*f;
}
const int maxn=500,mod=1000000007,inf=100000007;
bool cmin(int& a,int b){return a>b?a=b,true:false;}
bool cmax(int& a,int b){return a<b?a=b,true:false;}
int n;
double r;
struct node{
	double x,y,lef,rig;
}e[maxn];
bool cmp(node a,node b){return a.lef<b.lef;}
int main(){
	//freopen(FILE".in","r",stdin);
	//freopen(FILE".out","w",stdout);
	n=read(),r=read();
	up(i,1,n)e[i].x=read(),e[i].y=read(),e[i].rig=e[i].x+sqrt(r*r-e[i].y*e[i].y),e[i].lef=e[i].x-sqrt(r*r-e[i].y*e[i].y);
	sort(e+1,e+n+1,cmp);
	//up(i,1,n)printf("%.2lf %.2lf %.2lf %.2lf\n",e[i].x,e[i].y,e[i].lef,e[i].rig);
	int head=0,tail=0;
	node q[maxn];
	up(i,1,n){
		while(head<tail&&q[tail].rig>e[i].rig)tail--;
		q[++tail]=e[i];
	}
	n=tail-head+1;
	up(i,1,n)e[i]=q[i];
	//cout<<endl;
	//up(i,1,n)printf("%.2lf %.2lf %.2lf %.2lf\n",e[i].x,e[i].y,e[i].lef,e[i].rig);
	int top=0;
	int b[maxn];b[0]=-inf;
	up(i,1,n)if(b[top]<e[i].lef)b[++top]=e[i].rig;
	cout<<top<<endl;
	return 0;
}