Codeforces 8D Two Friends 三圆交集是否为非空集合 二分 旋转x轴

D. Two Friends

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Two neighbours, Alan and Bob, live in the city, where there are three buildings only: a cinema, a shop and the house, where they live. The rest is a big asphalt square.

Once they went to the cinema, and the film impressed them so deeply, that when they left the cinema, they did not want to stop discussing it.

Bob wants to get home, but Alan has to go to the shop first, and only then go home. So, they agreed to cover some distance together discussing the film (their common path might pass through the shop, or they might walk circles around the cinema together), and then to part each other's company and go each his own way. After they part, they will start thinking about their daily pursuits; and even if they meet again, they won't be able to go on with the discussion. Thus, Bob's path will be a continuous curve, having the cinema and the house as its ends. Alan's path — a continuous curve, going through the shop, and having the cinema and the house as its ends.

The film ended late, that's why the whole distance covered by Alan should not differ from the shortest one by more than t₁, and the distance covered by Bob should not differ from the shortest one by more than t₂.

Find the maximum distance that Alan and Bob will cover together, discussing the film.

Input

The first line contains two integers: t₁, t₂ (0 ≤ t₁, t₂ ≤ 100). The second line contains the cinema's coordinates, the third one — the house's, and the last line — the shop's.

All the coordinates are given in meters, are integer, and do not exceed 100 in absolute magnitude. No two given places are in the same building.

Output

In the only line output one number — the maximum distance that Alan and Bob will cover together, discussing the film. Output the answer accurate to not less than 4 decimal places.

Examples

input

0 2
0 0
4 0
-3 0

output

1.0000000000

input

0 0
0 0
2 0
1 0

output

2.0000000000

//#pragma comment(linker, "/STACK:102400000,102400000")
//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <bitset>
#include <iomanip>
#include <list>
#include <complex>
#include <stack>
#include <utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
const double eps = 1e-8;
const int INF = 1e9+7;
const ll inf =(1LL<<62) ;
const int MOD = 1e9 + 7;
const ll mod = (1LL<<32);
const int N =1e6+6;
const int M=100010;
const int maxn=1001;
#define mst(a) memset(a, 0, sizeof(a))
#define M_P(x,y) make_pair(x,y)
#define pi acos(-1.0)
#define in freopen("in.txt","r",stdin)
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
const int lowbit(int x) { return x&-x; }
int read(){ int v = 0, f = 1;char c =getchar();
while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}
while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();
return v*f;}
#define point complex<double>
double t1, t2;
point cinema, shop, house;

void readpoint(point &p)
{
  double x, y;
  scanf("%lf %lf", &x, &y);
  p = point(x, y);
}

bool inter(point a, double r_a, point b, double r_b, point c, double r_c) //以c为主圆求a b焦点判相交
{
  if (abs(c - a) <= r_a && abs(c - b) <= r_b) return true;
  //if (r_a + r_b < abs(b - a) || abs(b-a) + r_a < r_b || abs(b-a) + r_b < r_a) return false; 此处比较的精度不够会严重放大至二分的结果!!!!!
  b -= a;
  c -= a; //以a为原点
  point r = point(b.real() / abs(b), b.imag() / abs(b)); //将x轴正方向置为b
  b /= r;
  c /= r;
  double d = (r_a * r_a - r_b * r_b + abs(b) * abs(b)) / (2 * abs(b));
  double h = sqrt(max(r_a * r_a - d * d, 0.0));
  if (abs(h * h + (d - abs(b)) * (d - abs(b))) - r_b * r_b > eps) return false;//这行本地ac 测评ac
  if (abs(point(d, h) - c) <= r_c || abs(point(d, -h) - c) <= r_c) return true;
  return false;
}

bool check(point a, double r_a, point b, double r_b, point c, double r_c) //判断三圆是否相交
{
  if (r_a <= eps || r_b <= eps || r_c <= eps) return false; //有空集
  r_a = max(r_a, 0.0);
  r_b = max(r_b, 0.0);
  r_c = max(r_c, 0.0);
  if (inter(a, r_a, c, r_c, b, r_b)) return true;
  if (inter(b, r_b, a, r_a, c, r_c)) return true;
  if (inter(c, r_c, b, r_b, a, r_a)) return true;
  return false;
}

int main()
{
  scanf("%lf %lf", &t1, &t2);
  readpoint(cinema); //cinema
  readpoint(house); //house
  readpoint(shop); //shop

  if (abs(shop - cinema) + abs(house - shop) <= abs(cinema - house) + t2)//Alan <= Bob + t2
  {
    printf("%f\n", min( abs(cinema - house) + t2, abs(shop - cinema) + abs(house - shop) + t1));
  }
  else
  {
    double l, r, mid;
    l = 0;
    r = min( abs(cinema - house) + t2, abs(shop - cinema) + abs(house - shop) + t1);

    for(int i=1;i<=200;i++)
    {
      mid = (r + l) / 2;
      if (check(cinema, mid, shop, abs(shop - cinema) + t1 - mid, house, abs(house - cinema) + t2 - mid)){
        l = mid;
      }
      else {
        r = mid;
      }
    }
    printf("%f\n", l);
  }
  return 0;
}