hdu3124Arbiter(最小圆距离-扫描线)

时间:2023-03-09 17:06:44
hdu3124Arbiter(最小圆距离-扫描线)

链接

详解http://blog.sina.com.cn/s/blog_6e7b12310100qnex.html

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 100000
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
set<int>st;
struct point
{
double x,y,r;
point(double x = ,double y = ):x(x),y(y){}
int id;
}p[N],rank[N];
struct line
{
double po;
int id;
}lef[N],rig[N];
int rk[N],n,pp[N];
double mid;
typedef point pointt;
point operator -(point a,point b)
{
return point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
if(fabs(x)<eps) return ;
return x<?-:;
}
double dis(point a)
{
return sqrt(a.x*a.x+a.y*a.y*1.0);
}
int is_cross(int a,int b)
{
double dd = dis(rank[a]-rank[b]);
if(dcmp(dd-rank[a].r-rank[b].r-mid-mid)>) return ;
return ;
}
int judge(int a)
{
set<int>::iterator it=st.insert(a).first;//插入a后所在位置
if(it!=st.begin())
{
if(is_cross(a,*--it))
return ;
it++;
}
if(++it!=st.end())
{
if(is_cross(a,*it)) return ;
}
return ;
}
int cal()
{
st.clear();
int i = ,j = ;
while(i<=n||j<=n)
{
if(j==n+||(i!=n+&&dcmp(lef[i].po-mid-(rig[j].po+mid))<=))//如果当且i圆的最左端小于j圆的最右端 将i插进来
{
int ip = rk[lef[i].id];
if(judge(ip))
return ;
i++;
}
else
{
st.erase(rk[rig[j].id]);
j++;
}
}
return ;
}
double solve()
{
double low = 0.0,high = dis(p[]-p[])-p[].r-p[].r;
while(low+eps<high)//二分距离
{
mid = (high+low)/;
if(cal())
high = mid;
else low = mid;
}
return high+low;
}
bool cmp(line a,line b)
{
return a.po<b.po;
}
bool cmpp(point a,point b)
{
if(a.y==b.y) return a.x<b.x;
return a.y<b.y;
}
int main()
{
int t,i;
cin>>t;
while(t--)
{
scanf("%d",&n);
for(i = ; i <=n ;i++)
{
scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].r);
lef[i].po = p[i].x-p[i].r;//每个圆的最左端
lef[i].id = i;
rig[i].po = p[i].x+p[i].r;//每个圆的最右端
rig[i].id = i;
rank[i] = p[i];//按y坐标排序
rank[i].id = i;
}
sort(lef+,lef+n+,cmp);
sort(rig+,rig+n+,cmp);
sort(rank+,rank+n+,cmpp);
for(i = ; i <= n; i++)
{
rk[rank[i].id] = i;//每个圆按y坐标排序后位于第几
}
double ans = solve();
printf("%.6f\n",ans);
}
return ;
}