UVa 634 - Polygon

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题目：点与多边形关系判断。

分析：计算几何、点与多边形关系。在多边形外找到几个点与已知点连线，判断交点个数即可。会出现以下情况：穿过边，交点个数+1，穿过顶点交点个数+1（-1也可以，这里只是为了保证交点个数的奇偶性）。尝试多个点都满足交，点为奇数个则在多变形呢。如果已知点在多边形边上，则交点的奇偶性会发生变化，直接判断即可。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cmath>using namespace std;//点结构 typedef struct pnode{double x,y;pnode( double a, double b ){x = a;y = b;}pnode(){}}point;point D[1005];//线结构 typedef struct lnode{double x,y,dx,dy;lnode( point a, point b ){x = a.x;y = a.y;dx = b.x-a.x;dy = b.y-a.y;}lnode(){}}line;//点在线段上 bool on( point p, line l ){if ( l.dx*(p.y-l.y)-l.dy*(p.x-l.x) == 0 )if ( (p.x-l.x)*(p.x-l.x-l.dx) <= 0 )if ( (p.y-l.y)*(p.y-l.y-l.dy) <= 0 )return true;return false;}//线段相交 bool cross( line a, line b ){double t1 = a.dx*(b.y-a.y)-a.dy*(b.x-a.x);double t2 = a.dx*(b.y+b.dy-a.y)-a.dy*(b.x+b.dx-a.x);double t3 = b.dx*(a.y-b.y)-b.dy*(a.x-b.x);double t4 = b.dx*(a.y+a.dy-b.y)-b.dy*(a.x+a.dx-b.x);return (t1*t2 <= 0)&&(t3*t4 <= 0);}//点在多边形内bool in( point p, point* P, int n ){double d[4][2] = {-1e6,-1e6,-1e6,1e6,1e6,-1e6,1e6,1e6};for ( int t = 0 ; t < 4 ; ++ t ) {line s1 = line( p, point( d[t][0], d[t][1] ) );int  count = 0;for ( int i = 0 ; i < n ; ++ i ) {line s2 = line( P[i], P[i+1] );if ( on( p, s2 ) ) return true;if ( cross( s1, s2 ) ) count ++;if ( on( P[i], s1 ) ) count ++;}if ( count%2 == 0 ) return false; }return true;}int main(){int n;while ( scanf("%d",&n) && n ) {for ( int i = 0 ; i <= n ; ++ i )scanf("%lf%lf",&D[i].x,&D[i].y);D[n+1] = D[n]; D[n+0] = D[0];if ( in( D[n+1], D, n ) )printf("T\n");else printf("F\n");}return 0;}