ZOJ 3570 Lott's Seal 计算几何

题目大意是

有一些点，然后我们要用一个六角星形将任意这些点连成的直线覆盖。 并且这些点构成的凸包面积必须满足小于某个值

六角星形的中心点和半径已经给定了。

就是一个判定问题了。

首先要判断所有点是否都在六角星形内

我们观察这个形状，发现是两个三角形组成的图形。

那么只需判断某个点是否在某个三角形内即可

这里就用到叉积就行了。

然后对所有点求个凸包。

求个面积

然后看这些凸包的边是否在六角星形内

观察图形，可以发现其由12个短线段构成边界

那么看边是否在六角星形内。只需判断某个边是否与这些边界规范相交即可

代码如下。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <cmath>#include <map>#include <sstream>#include <queue>#include <vector>#define MAXN 111111#define MAXM 211111#define eps 1e-8#define INF 1000000001using namespace std;int dblcmp(double d){    if (fabs(d) < eps) return 0;    return d > eps ? 1 : -1;}struct point{    double x, y;    point(){}    point(double _x, double _y):    x(_x), y(_y){};    void input()    {        scanf("%lf%lf",&x, &y);    }    double dot(point p)    {        return x * p.x + y * p.y;    }    double distance(point p)    {        return hypot(x - p.x, y - p.y);    }    point sub(point p)    {        return point(x - p.x, y - p.y);    }    double det(point p)    {        return x * p.y - y * p.x;    }    bool operator < (point a)const    {        return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;    }}p[MAXN], hg[15];struct line{    point a, b;    line(){}    line(point _a, point _b){ a = _a; b = _b;}    int segcrossseg(line v)    {        int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));        int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));        int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));        int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));        if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;        return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <=0 ||                d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <=0 ||                d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0 ||                d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);    }}seg[13], tri[2][3];struct polygon{    int n;    point p[MAXN];    line l[MAXN];    double area;    void input()    {        for(int i = 0; i < n; i++) p[i].input();    }    void getline()    {        for(int i = 0; i < n; i++)            l[i] = line(p[i], p[(i + 1) % n]);    }    void getarea()    {        area = 0;        int a = 1, b = 2;        while(b <= n - 1)        {            area += p[a].sub(p[0]).det(p[b].sub(p[0]));            a++;            b++;        }        area = fabs(area) / 2;    }}convex;bool conpoint(point p[],int n){    for (int i = 1; i < n; i++)        if (dblcmp(p[i].x - p[0].x) != 0 ||                dblcmp(p[i].y - p[0].y) != 0)            return false;    return true;}bool conline(point p[],int n){    for (int i = 2; i < n; i++)        if (dblcmp(p[1].sub(p[0]).det(p[i].sub(p[0])))  != 0)   return false;    return true;}void getconvex(point p[], int n, point res[], int& resn){    resn = 0;    if (conpoint(p, n))    {        res[resn++] = p[0];        return;    }    sort(p, p + n);    if (conline(p,n))    {        res[resn++] = p[0];        res[resn++] = p[n - 1];        return;    }    for (int i = 0; i < n;)        if (resn < 2 || dblcmp(res[resn - 1].sub(res[resn - 2]).det(p[i].sub(res[resn - 1]))) > 0)            res[resn++] = p[i++];        else --resn;    int top = resn - 1;    for (int i = n - 2; i >= 0;)        if (resn < top + 2 || dblcmp(res[resn - 1].sub(res[resn - 2]).det(p[i].sub(res[resn - 1]))) > 0)            res[resn++] = p[i--];        else --resn;    resn--;}int n;bool intriangle(point x){    int tmp = 2;    for(int i = 0; i < 2; i++)    {        for(int j = 0; j < 3; j++)            if(dblcmp(tri[i][j].a.sub(x).det(tri[i][j].b.sub(x))) > 0)            {                tmp--;                break;            }    }    return tmp > 0;}int main(){    double sx, sy, S, R;    while(scanf("%lf%lf", &sx, &sy) != EOF)    {        scanf("%d", &n);        for(int i = 0; i < n; i++) p[i].input();        getconvex(p, n, convex.p, convex.n);        convex.getline();        convex.getarea();        scanf("%lf%lf", &R, &S);        double sthree = sqrt(3.0);        hg[0] = point(sx, sy + sthree * R);        hg[1] = point(sx + R / 2, sy + sthree * R / 2);        hg[2] = point(sx + R * 3.0 / 2, sy + sthree * R / 2);        hg[3] = point(sx + R, sy);        hg[4] = point(sx + R * 3.0 / 2, sy - sthree * R / 2);        hg[5] = point(sx + R / 2, sy - sthree * R / 2);        hg[6] = point(sx, sy - sthree * R);        hg[7] = point(sx - R / 2, sy - sthree * R / 2);        hg[8] = point(sx - R * 3.0 / 2, sy - sthree * R / 2);        hg[9] = point(sx - R, sy);        hg[10] = point(sx - R * 3.0 / 2, sy + sthree * R / 2);        hg[11] = point(sx - R / 2, sy + sthree * R / 2);        tri[0][0] = line(hg[0], hg[4]);        tri[0][1] = line(hg[4], hg[8]);        tri[0][2] = line(hg[8], hg[0]);        tri[1][0] = line(hg[2], hg[6]);        tri[1][1] = line(hg[6], hg[10]);        tri[1][2] = line(hg[10], hg[2]);        for(int i = 0; i < 12; i++) seg[i] = line(hg[i], hg[(i + 1) % 12]);        int flag = 1;        for(int i = 0; i < n; i++)        {            if(!intriangle(p[i]))            {                flag = 0;                break;            }        }        for(int i = 0; i < 12; i++)        {            if(flag == 0) break;            for(int j = 0; j < convex.n; j++)                if(seg[i].segcrossseg(convex.l[j]) == 2)                {                    flag = 0;                    break;                }        }        if(flag)        {            if(dblcmp(3.0 * sthree * R * R - convex.area - S) > 0) puts("Succeeded.");            else puts("Failed.");        }        else puts("Failed.");    }    return 0;}