Eligible Segments (CF 1588 E)

题目链接
题目大意：
给定 n 个点，问从中能找出多少条线段，使剩余n-2个点到该线段的距离均不大于R。
前置知识：
点a到线段bc的距离，可以转换为点a到射线bc和射线cb距离的最大值。
解题思路：
要使线段p[i]p[j]到剩余n-2个点的距离均小于R,即满足射线p[i][j]到剩余n-2个点的距离小于等于R && 射线p[j]p[i]到剩余n-2个点的距离小于等于R。
而对于一个点来说，能满足其出发的射线到剩余n-1个点距离都小于等于R即求该点到剩余n-1个以R为半径的圆的两条切线的角度之交。
求出合法区间后就只需确认对于该点满足的区间内的点进行标记，最后符合答案的情况即：

if(mp[i][j] && mp[j][i]) Ans ++;
//即设任意一点到该线段p[i][j]的距离为d
//点到射线[p[i], p[j]) 的距离为d1
//点到射线[p[j], p[i]) 的距离为d2
//d = max(d1, d2) , 因为 d1 <= R && d2 <= R
// 因此满足d <= R

代码如下：

#include <bits/stdc++.h>
#define In inline
#define pi (atan(1.0)*4)
#define enter puts("")
#define MaxN 0x3f3f3f3f
#define MinN 0xc0c0c0c0
#define pb push_back
#define bug(x) cerr<<#x<<'='<<x<<' '
#define ALL(a) a.begin(), a.end()
#define rep(i,a,b) for(int i=(a);i <=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define mset(a,b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define eps 1e-6
#define buff ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
template<typename T>
T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<typename T>
T lcm(T a,T b){return a*b/gcd(a,b);}
typedef long long ll;
typedef pair<int,int> PII;
const int mod = 998244353;
const int MOD = 1e9+7;
In ll read()
{ll ans = 0;char ch = getchar(), las = ' ';while(!isdigit(ch)) las = ch, ch = getchar();while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();if(las == '-') ans = -ans;return ans;
}
In void write(ll x)
{if(x < 0) x = -x, putchar('-');if(x >= 10)   write(x / 10);putchar(x%10 + '0');
}struct Point{double x, y;Point(){}Point(double x, double y):x(x), y(y){}void input(){scanf("%lf%lf", &x, &y);}Point operator + (Point b){return Point(x+b.x, y+b.y);}Point operator - (Point b){return Point(x-b.x, y-b.y);}double angle(Point b){return atan2(b.y-y, b.x-x);}double dis(Point b){return sqrt((x-b.x)*(x-b.x)+(y-b.y)*(y-b.y));}
};
const int N = 3100;
int mp[N][N];
Point p[N];void run_case(){int n = read(), Rad = read();rep(i, 1, n) p[i].input();rep(i, 1, n){double L = -pi, R = pi;bool ck = 1, flag = 1;rep(j, 1, n){if(i != j){double ang = p[i].angle(p[j]), r, l, d = p[i].dis(p[j]);if(d <= Rad) continue;//此时贡献限制区间为[-pi, pi], passdouble a = asin(Rad*1.0/d);r = ang + a, l = ang - a;if(ck) L = l, R = r, ck = 0;else{if(l > R + eps)    l -= pi*2, r -= pi*2;if(r < L + eps) l += pi*2, r += pi*2;L = max(L, l), R = min(R, r);}if(L > R + eps){flag = 0;break;}//合法区间不存在}}if(flag){rep(j, 1, n){if(i == j) continue;double ang = p[i].angle(p[j]);if(ang > R + eps) ang -= 2 * pi;if(ang < L - eps) ang += 2 * pi;if(L - eps < ang && ang < R + eps) mp[i][j] = 1;}}}int Ans = 0;rep(i, 1, n) rep(j, i+1, n){if(mp[i][j] && mp[j][i]) Ans ++;}write(Ans);enter;
}int main()
{//  freopen("C:\\Users\\MARX HE\\Desktop\\input.txt","r",stdin);
//  freopen("C:\\Users\\MARX HE\\Desktop\\output.txt","w",stdout);int _ = 1;
//    int _ = read();while(_ --){run_case();}
}

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