[SCOI2015]小凸想跑步——半平面交

题面

     bzoj

解析

    设凸包的0，1号点为A(x1,y1)，B(x2,y2),在剩余点中任取两个相邻点C(x3,y3)，D(x4,y4)，满足点D对应的点的编号大于点C对应的点的编号，再设凸包中可选点为P(x,y)

由叉积求面积公式得：

（x1-x) * (y2-y) - (x2-x) * (y1-y) < (x3-x) * (y4-y) - (x4-x) * (y3-y)

  （这里求面积不用加绝对值，逆时针计算叉积，这里一定为正）

 化简得：

(y1-y2-y3+y4) * x + (-x1+x2+x3-x4) * y + x1*y2 - x2 * y1 - x3 * y4 + x4 * y3 < 0

令A =  y1-y2-y3+y4,   B = -x1+x2+x3-x4,   C = x1*y2 - x2 * y1 - x3 * y4 + x4 * y3

原不等式变为 Ax + By + C < 0

可以联想到线性规划， 即半平面交

再考虑，如何将不等式转化为向量，并且满足向量的左半平面是可行域

这个可以自己画一画，发现无论A，B为正或负，当该直线的方向向量为( -B, A)时，满足要求

向量起点为便于计算可以取直线与任一坐标轴交点，注意特判直线与该坐标轴是否有交点，没有则取直线与另一坐标轴交点

把原凸包的所有边和用该不等式构造出的边加入坐标系，跑一下半平面交的模板就可以了

代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 200005 ;
const double eps = 1e-13;

int n, tot, L, R;
struct point{
    double xx, yy;
}p[maxn], a[maxn];

struct line{
    double deg;
    point u, v;
    void Mak(int x,int y)
    {
        u = p[x];
        v = p[y];
        deg = atan2(p[y].yy - p[x].yy, p[y].xx - p[x].xx);
    }
}xl[maxn], qu[maxn];

double Cp(point x, point y)
{
    return (double)(x.xx * y.yy) - (x.yy  * y.xx);
}

bool cmp(line x, line y)
{
    if(fabs(x.deg - y.deg) >= eps)
        return x.deg < y.deg;
    point p = (point){y.v.xx - x.u.xx, y.v.yy - x.u.yy}, q = (point){x.v.xx - x.u.xx, x.v.yy - x.u.yy};
    return Cp(p, q) < 0;
}

point Getpoint(line a, line b)
{
    double a1 = a.u.yy - a.v.yy, b1 = a.v.xx - a.u.xx, c1 = a.u.xx * a.v.yy - a.v.xx * a.u.yy;
    double a2 = b.u.yy - b.v.yy, b2 = b.v.xx - b.u.xx, c2 = b.u.xx * b.v.yy - b.v.xx * b.u.yy;
    return (point){(c1*b2-c2*b1)/(a2*b1-a1*b2), (a2*c1-a1*c2)/(a1*b2-a2*b1)};
}

bool check(point x, line y)
{
    double z = Cp((point){y.v.xx - y.u.xx, y.v.yy - y.u.yy}, (point){x.xx - y.u.xx, x.yy - y.u.yy});
    return z < 0.0;
}

int deq[maxn];

void SI()
{
    sort(xl + 1, xl + tot + 1, cmp);
    int cnt = 0;
    for(int i = 1 ; i < tot ; ++i)
        if(fabs(xl[i].deg - xl[i+1].deg) >= eps)
            qu[++cnt] = xl[i];
    qu[++cnt] = xl[tot];
    L = 1;R = 2;
    deq[1] = 1;deq[2] = 2;
    for(int i = 3; i <= cnt ; ++i)
    {
        while(L < R && check(Getpoint(qu[deq[R]], qu[deq[R-1]]), qu[i]))    R--;
        while(L < R && check(Getpoint(qu[deq[L]], qu[deq[L+1]]), qu[i]))    L++;
        deq[++R] = i;
    }
    while(L < R && check(Getpoint(qu[deq[R]], qu[deq[R-1]]), qu[deq[L]]))    R--;
    while(L < R && check(Getpoint(qu[deq[L]], qu[deq[L+1]]), qu[deq[R]]))    L++;
}

double Size_big()
{
    double ret = 0.0;
    for(int i = 1; i <= n ;++i)
        ret += Cp(p[i], p[i+1]);
    return fabs(ret);
}

double Size_small()
{
    double ret = 0.0;
    int res = 0;
    deq[R + 1] = deq[L];
    for(int i = L ; i <= R; ++i)
        a[++res] = Getpoint(qu[deq[i]], qu[deq[i + 1]]);
    for(int i = 1; i < res ; ++i)
        ret += Cp(a[i], a[i + 1]);
    ret += Cp(a[res], a[1]);
    return fabs(ret);
}

int main()
{
    scanf("%d",&n);
    tot = n;
    for(int i = 1; i <= n ; ++i)
        scanf("%lf%lf", &p[i].xx, &p[i].yy);
    p[n + 1] = p[1];
    double S = Size_big();
    for(int i = 1; i <= n ; ++i)
        xl[i].Mak(i, i + 1);
    for(int i = 2 ; i <= n ; ++i)
    {
        double A = p[1].yy - p[i].yy - p[2].yy + p[i + 1].yy;
        double B = p[2].xx - p[i + 1].xx - p[1].xx + p[i].xx;
        double C = Cp (p[1], p[2]) - Cp (p[i], p[i + 1]);
        xl[++tot].u.xx = B ? 0.0 : -C / A;
        xl[tot].u.yy = B ? -C / B : 0.0;
        xl[tot].v = (point) {xl[tot].u.xx - B, xl[tot].u.yy + A};
        xl[tot].deg = atan2(xl[tot].v.yy - xl[tot].u.yy, xl[tot].v.xx - xl[tot].u.xx);
    }
    SI();
    double s = Size_small();

    printf("%.4lf", s / S);
    return 0;
}