XMOJ 1133: 膜拜大牛 计算几何/两圆相交

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff;   //无限大
const int inf=0x3f3f3f3f;
/*
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
*/
//**************************************************************************************


const double  eps = 1e-9;
const double pi = acos(-1.0);
const int N = 555;
    
struct point
{
    double x,y;
    int nb;
}p[N];
    
int n,r;
    
struct alpha
{
    double v;
    int nb;
    int flag;
    bool friend operator <(const alpha &a,const alpha &b)
    {   
        if(abs(a.v-b.v)<=eps)
            return a.flag>b.flag;
        else
            return a.v < b.v;
    }
}alp[N * 2];
    
double dis(point a, point b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
    
void solve(double R)
{
    int ans = 0;
    double theta,phi,dist;
    int i,j;
    for( i = 0;i < n;i++)
    {
        int k = 0;
        for( j = 0;j < n;j++)
        {
            if(i == j) continue;
            dist = dis(p[i],p[j]);
            if(dist - 2.0 * R > eps)//判断是否有交点
                continue;
            theta = atan2(p[j].y - p[i].y, p[j].x - p[i].x);//这条直线的斜率
            if(theta < eps)
                theta += 2 * pi;
            phi = acos(dist / (2.0 * R));
            alp[k].v = theta - phi + 2 * pi;
            alp[k].flag = 1;
            alp[k].nb=p[j].nb;
  
            alp[k + 1].v = theta + phi + 2 * pi;
            alp[k + 1].flag = 0;
            alp[k+1].nb=p[j].nb;
            k += 2;
        }
        sort(alp,alp + k);
        int sum = p[i].nb;
        ans=max(sum,p[i].nb);
        for( j = 0;j < k;j++)
        {
            if(alp[j].flag)
                sum +=alp[j].nb;
            else
                sum -=alp[j].nb;
            if(sum > ans)
                ans = sum;
        }
    }
    printf("%d\n",ans );
}
    
int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            int x,y;
            int nb;
            scanf("%d %d %d",&x,&y,&nb);
            p[i].x=double(x);
            p[i].y=double(y);
            p[i].nb=nb;
        }
        scanf("%d",&r);
        solve(double(r));
    }
    return 0;
}