poj1584

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。

分析：计算几何

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

首先，若点是顺时针则reverse()改为逆时针，reverse函数就是用来把数组反向的。然后利用每3个相邻点组成的两条向量的叉积来判断，都应大于等于零。然后，判断是否在内部，利用钉子点和多边形每两个相邻点，组成两个向量。判断叉积是否全都大于0（全为逆时针）。再就是判断点到直线的距离，利用三角形面积除以底边，面积用叉积求。

View Code

#include < iostream >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cmath >
#include < algorithm >
using namespace std;

#define maxn 2000
#define eps 10E-9

struct 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()
{
// freopen("t.txt", "r", stdin);
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 ;
}