POJ 1584 A Round Peg in a Ground Hole（点到直线距离，圆与多边形相交，多边形是否为凸）

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题意：给出一个多边形和一个圆，问是否是凸多边形，若是则再问圆是否在凸多边形内部。

分3步：

1、判断是否是凸多边形

2、判断点是否在多边形内部

3、判断点到各边的距离是否大于等于半径

上代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>using namespace std;#define maxn 2000#define eps 10E-8struct Point{    double x, y;    Point operator-(const Point &a) const    {        Point ret;        ret.x = x - a.x;        ret.y = y - a.y;        return ret;    }} point[maxn], peg;double pegr;int n;int dblcmp(double a){    if (fabs(a) < eps)        return 0;    return a >0?1 : -1;}double xmult(const Point &a, const Point &b){    return a.x * b.y - b.x * a.y;}void input(){    scanf("%lf%lf%lf", &pegr, &peg.x, &peg.y);    for (int i =0; i < n; i++)        scanf("%lf%lf", &point[i].x, &point[i].y);    int t =0;    int i =0;    while (i < n && t ==0)    {        t = dblcmp(xmult(point[(i +1)%n] - point[i], point[(i +2)%n] - point[i]));        i++;    }    if (t <0)        reverse(point, point + n);}bool convex(){    for (int i =0; i < n; i++)        if (dblcmp(xmult(point[(i +1)%n] - point[i], point[(i +2)%n] - point[(i +1)%n]))    <0)            return false;    return true;}bool inconvex(){    for (int i =0; i < n; i++)        if (dblcmp(xmult(point[(i +1)%n] - point[i], peg - point[(i +1)%n])) <=0)            return false;    return true;}double dist(const Point &a, const Point &b){    Point p;    p = a - b;    return sqrt(p.x * p.x + p.y * p.y);}bool ok(){    for (int i =0; i < n; i++)        if (dblcmp(abs(xmult(peg - point[i], point[(i +1)%n] - point[i]))/dist(point[i], point[(i +1)%n]) - pegr) <0)            return false;    return true;}int main(){    while (scanf("%d", &n) != EOF)    {        if (n<3)            break;        input();        if (!convex())            printf("HOLE IS ILL-FORMED\n");        else if (!inconvex()||!ok())            printf("PEG WILL NOT FIT\n");        else            printf("PEG WILL FIT\n");    }    return 0;}

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