POJ 1584 计算几何 凸包

链接：

http://poj.org/problem?id=1584

题意：

按照顺时针或逆时针方向输入一个n边形的顶点坐标集，先判断这个n边形是否为凸包。

再给定一个圆形（圆心坐标和半径），判断这个圆是否完全在n变形内部。

题解：

1、判断凸包convex()：

   由于点集已经按某个时针方向有序，因此可以先定义一个方向系数dir=0

   两两枚举n边形的边，用叉积判断这两条边的转向（右螺旋或左螺旋），由于存在散点共线的情况，因此当且仅当叉积的值t第一次不为0时，dir=t，dir的值此后不再改变。（dir>0 则为右螺旋逆时针，dir<0则为左螺旋顺时针）

   此后继续枚举剩下的边，只要判断dir*t>=0即可，当存在一个dir*t<0的边，说明这是凹多边形，就不是凸包了。

2、判断圆心在不在凸包内contain()

3、当圆心在凸包内时，判断距离是否都大于半径fit()

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define mp make_pair
#define lson l,m,rt<<1  
#define rson m+1,r,rt<<1|1 
typedef long long ll;
typedef vector<int> VI;
typedef pair<int, int> PII;
const ll MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1010;
// head

const double eps = 1e-8;
int cmp(double x) {
    if (fabs(x) < eps) return 0;
    if (x > 0) return 1;
    return -1;
}

const double pi = acos(-1);
inline double sqr(double x) {
    return x*x;
}
struct point {
    double x, y;
    point() {}
    point(double a, double b) :x(a), y(b) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
    friend point operator+(const point &a, const point &b) {
        return point(a.x + b.x, a.y + b.y);
    }
    friend point operator-(const point &a, const point &b) {
        return point(a.x - b.x, a.y - b.y);
    }
    friend point operator*(const double &a, const point &b) {
        return point(a*b.x, a*b.y);
    }
    friend point operator/(const point &a, const double &b) {
        return point(a.x / b, a.y / b);
    }
    double norm() {
        return sqrt(sqr(x) + sqr(y));
    }
};
double det(point a, point b) {
    return a.x*b.y - a.y*b.x;
}
double dot(point a, point b) {
    return a.x*b.x + a.y*b.y;
}
double dist(point a, point b) {
    return (a - b).norm();
}

struct line {
    point a, b;
    line() {}
    line(point x, point y) :a(x), b(y) {}
};
double dis_point_segment(point p, point s, point t) {
    if (cmp(dot(p - s, t - s)) < 0) return (p - s).norm();
    if (cmp(dot(p - t, s - t)) < 0) return (p - t).norm();
    return fabs(det(s - p, t - p) / dist(s, t));
}
bool point_on_segment(point p, point s, point t) {
    return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0;
}
bool parallel(line a, line b) {
    return !cmp(det(a.a - a.b, b.a - b.b));
}
bool line_make_point(line a, line b,point &res) {
    if (parallel(a, b)) return false;
    double s1 = det(a.a - b.a, b.b - b.a);
    double s2 = det(a.b - b.a, b.b - b.a);
    res = (s1*a.b - s2*a.a) / (s1 - s2);
    return true;
}

int n;
double r;
point o;
point *p;

bool convex() {
    int dir = 0;
    rep(i, 0, n) {
        int t = cmp(det(p[i + 1] - p[i], p[i + 2] - p[i + 1]));
        if (!dir) dir = t;
        if (dir*t < 0) return false;
    }
    return true;
}

bool contain() {
    int sign = 0;
    rep(i, 0, n) {
        int x = cmp(det(p[i] - o, p[i + 1] - o));
        if (x) {
            if (sign && sign != x) return false;
            else sign = x;
        }
    }
    return true;
}

bool fit() {
    rep(i, 0, n) {
        int k = cmp(dis_point_segment(o, p[i], p[i + 1]) - r);
        if (k < 0) return false;
    }
    return true;
}

int main() {
    while (cin >> n && n != 1) {
        cin >> r;
        o.input();
        p = new point[n + 2];
        rep(i, 1, n + 1) p[i].input();
        p[0] = p[n];
        p[n + 1] = p[1];
        if(!convex()) cout << "HOLE IS ILL-FORMED" << endl;
        else {
            bool fg1 = contain();
            bool fg2 = fit();
            if (fg1 && fg2) cout << "PEG WILL FIT" << endl;
            else cout << "PEG WILL NOT FIT" << endl;
        }
        delete p;
    }
    return 0;
}