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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#313796#8082. Minimum Euclidean DistanceEndPBWA 0ms6976kbC++176.2kb2024-01-24 23:49:252024-01-24 23:49:26

Judging History

你现在查看的是最新测评结果

  • [2024-01-24 23:49:26]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:6976kb
  • [2024-01-24 23:49:25]
  • 提交

answer

#include<bits/stdc++.h>

using namespace std;
#define endl '\n'
#define ll long long
#define ld long double
#define Vector Point
#define S(x) (x)*(x)
#define N 100010

const int Kep = 10;
const ld eps = 1e-9, pi = acosl(-1.0), Mn = powl(10.0, -Kep);

inline int sign(ld a) { return a < -eps ? -1 : (a > eps ? 1 : 0); }

inline ld Abs(ld a) { return a * sign(a); }//取绝对值
inline void Out(long double a, int k = Kep) {
    if (-Mn < a && a < Mn) a = 0;
    cout << fixed << setprecision(Kep) << a;//<iomanip>
}

struct Point {
    ld x, y;

    Point(ld a = 0, ld b = 0) { x = a, y = b; }

    inline void in() { cin >> x >> y; }

    inline void out() {
        Out(x);
        cout << " ";
        Out(y);
        cout << endl;
    }
};

inline ld Dot(Vector a, Vector b) { return a.x * b.x + a.y * b.y; }//【点积】
inline ld Cro(Vector a, Vector b) { return a.x * b.y - a.y * b.x; }//【叉积】
inline ld Len(Vector a) { return sqrtl(Dot(a, a)); }//【模长】
inline ld Angle(Vector a, Vector b) { return acosl(Dot(a, b) / Len(a) / Len(b)); }//【两向量夹角】
inline Vector Normal(Vector a) { return Vector(-a.y, a.x); }//【法向量】
inline Vector operator+(Vector a, Vector b) { return Vector(a.x + b.x, a.y + b.y); }

inline Vector operator-(Vector a, Vector b) { return Vector(a.x - b.x, a.y - b.y); }

inline Vector operator*(Vector a, ld b) { return Vector(a.x * b, a.y * b); }

bool operator<(const Point &a, const Point &b) { return Abs(a.x - b.x) < eps ? a.y < b.y : a.x < b.x; }

inline bool operator==(Point a, Point b) { return !sign(a.x - b.x) && !sign(a.y - b.y); }//两点坐标重合则相等
inline bool same_dir(Vector a, Vector b) { return sign(Cro(b, a)) == 0 && sign(Dot(b, a)) > 0; }//判断两向量是否同向,退化为点后会返回0
inline int pan_PS(Point p, Point a, Point b) {//【判断点P是否在线段AB上】
    return !sign(Cro(p - a, b - a)) && sign(Dot(p - a, p - b)) <= 0;//做法一
    //  return !dcmp(Cro(p-a,b-a))&&dcmp(min(a.x,b.x)-p.x)<=0&&dcmp(p.x-max(a.x,b.x))<=0&&dcmp(min(a.y,b.y)-p.y)<=0&&dcmp(p.y-max(a.y,b.y))<=0;//做法二
    //PA,AB共线且P在AB之间(其实也可以用len(p-a)+len(p-b)==len(a-b)判断,但是精度损失较大)
}

inline int pan_PL(Point p, Point a, Point b) {//【判断点P是否在直线AB上】
    return !sign(Cro(p - a, b - a));//PA,AB共线
}

inline ld dis_PL(Point p, Point a, Point b) {
    if (a == b)return Len(p - a);//AB重合
    Vector x = p - a, y = p - b, z = b - a;
    return Abs(Cro(x, z) / Len(z));//面积除以底边长
}

inline ld dis_PS(Point p, Point a, Point b) {//【点P到线段AB距离】
    if (a == b)return Len(p - a);//AB重合
    Vector x = p - a, y = p - b, z = b - a;
    if (sign(Dot(x, z)) < 0)return Len(x);//P距离A更近
    if (sign(Dot(y, z)) > 0)return Len(y);//P距离B更近
    return Abs(Cro(x, z) / Len(z));//面积除以底边长
}

inline Point FootPoint(Point p, Point a, Point b) {//点P到直线AB的垂足
    Vector x = p - a, y = p - b, z = b - a;
    ld len1 = Dot(x, z) / Len(z), len2 = -1.0 * Dot(y, z) / Len(z);//分别计算AP,BP在AB,BA上的投影
    return a + z * (len1 / (len1 + len2));//点A加上向量AF
}

inline Point poi_PS(Point p, Point a, Point b) {//【点P到线段AB的最近位置】
    if (a == b)return Len(p - a);//AB重合
    Vector x = p - a, y = p - b, z = b - a;
    if (sign(Dot(x, z)) < 0)return a;//P距离A更近
    if (sign(Dot(y, z)) > 0)return b;//P距离B更近
    return FootPoint(p, a, b);//返回垂足
}

inline Point Symmetry_PL(Point p, Point a, Point b) {//【点P关于直线AB的对称点】
    return p + (FootPoint(p, a, b) - p) * 2;//将PF延长一倍即可
}

inline Point cross_LL(Point a, Point b, Point c, Point d) {//【两直线AB,CD的交点】
    Vector x = b - a, y = d - c, z = a - c;
    return a + x * (Cro(y, z) / Cro(x, y));//点A加上向量AF
}

inline int pan_cross_LS(Point a, Point b, Point c, Point d) {//【判断直线AB与线段CD是否相交】
    return pan_PL(cross_LL(a, b, c, d), c, d);//直线AB与直线CD的交点在线段CD上
}

inline int pan_cross_SS(Point a, Point b, Point c, Point d) {//【判断两线段AB,CD是否相交】
    ld c1 = Cro(b - a, c - a), c2 = Cro(b - a, d - a);
    ld d1 = Cro(d - c, a - c), d2 = Cro(d - c, b - c);
    return sign(c1) * sign(c2) < 0 && sign(d1) * sign(d2) < 0;//分别在两侧
}

inline int PIP(Point *P, int n, Point a) {//【射线法】判断点A是否在任意多边形Poly以内 O(n)
    int cnt = 0;
    ld tmp;
    for (int i = 1; i <= n; ++i) {
        int j = i < n ? i + 1 : 1;
        if (pan_PL(a, P[i], P[j]))return 2;//点在多边形上
        if (a.y >= min(P[i].y, P[j].y) && a.y < max(P[i].y, P[j].y))//纵坐标在该线段两端点之间
            tmp = P[i].x + (a.y - P[i].y) / (P[j].y - P[i].y) * (P[j].x - P[i].x), cnt += sign(tmp - a.x) > 0;//交点在A右方
    }
    return cnt & 1;//穿过奇数次则在多边形以内
}

inline int judge_PP(Point *A, int n, Point *B, int m) {//【判断多边形A与多边形B是否相离】
    for (int i1 = 1; i1 <= n; ++i1) {
        int j1 = i1 < n ? i1 + 1 : 1;
        for (int i2 = 1; i2 <= m; ++i2) {
            int j2 = i2 < m ? i2 + 1 : 1;
            if (pan_cross_SS(A[i1], A[j1], B[i2], B[j2]))return 0;//两线段相交
            if (PIP(B, m, A[i1]) || PIP(A, n, B[i2]))return 0;//点包含在内
        }
    }
    return 1;
}

inline bool pan_ALR(Point A, Point L, Point R) {//判断AL在AR严格左侧
    return sign(Cro(L - A, R - A)) < 0;
}

int n, m;
Point p[N], A, B, O;
ld r = 0;

void solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        p[i].in();
    }
    for (int i = 1; i <= m; ++i) {
        A.in(), B.in();
        O = A + B;
        O.x /= 2, O.y /= 2;
        r = Len(A - B) / 2;
        if (PIP(p, n, O)) {
            printf("%.10Lf\n", r * r / 2);
            continue;
        }
        ld mn = 1e100;
        for (int j = 1; j <= n; ++j) {
            mn = min(mn, dis_PS(O, p[j], p[j == n ? 1 : n + 1]));
        }
        mn = S(r) / 2 + S(mn);
        printf("%.10Lf\n", mn);
    }
}

signed main() {
    int T = 1;
//    cin >> T;
    while (T--) solve();
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 6976kb

input:

4 3
0 0
1 0
1 1
0 1
0 0 1 1
1 1 2 2
1 1 2 3

output:

0.2500000000
0.7500000000
1.8750000000

result:

ok Your answer is acceptable!^ ^

Test #2:

score: -100
Wrong Answer
time: 0ms
memory: 6976kb

input:

48 10
-30 0
-29 -4
-28 -7
-27 -9
-25 -12
-22 -16
-21 -17
-17 -20
-14 -22
-12 -23
-9 -24
-5 -25
-4 -25
0 -24
3 -23
5 -22
8 -20
12 -17
13 -16
16 -12
18 -9
19 -7
20 -4
21 0
21 1
20 5
19 8
18 10
16 13
13 17
12 18
8 21
5 23
3 24
0 25
-4 26
-5 26
-9 25
-12 24
-14 23
-17 21
-21 18
-22 17
-25 13
-27 10
-28 ...

output:

589.5000000000
52.0000000000
1051.2500000000
66.6250000000
174.3750000000
562.8750000000
272.8750000000
287.8750000000
689.6250000000
436.2500000000

result:

wrong answer Except 51.470588235300, but found 52.000000000000!QAQ