/**//*
判断点q是否在多边形内的一种方法，过q作水平射线L，
如果L与多边形P的边界不相交，则q在P的外部。否则，
L和P的边界相交，具体地说，交点个数为奇（偶）数
时，点q在P的内（外）部。 */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MaxNode 50
#define INF 999999999
typedef struct TPoint
{
double x;
double y;
}TPoiont;
typedef struct TSegment
{
TPoint p1;
TPoint p2;
}TSegment;
typedef struct TPolygon
{
TPoint point[MaxNode];
int n;
}TPolygon;
double multi(TPoint p1, TPoint p2, TPoint p0)
{
//求矢量[p0, p1], [p0, p2]的叉积
//p0是顶点
return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
//若结果等于0，则这三点共线
//若结果大于0，则p0p2在p0p1的逆时针方向
//若结果小于0，则p0p2在p0p1的顺时针方向
}
double max(double x, double y)
{
//比较两个数的大小，返回大的数
if(x > y) return x;
else return y;
}
double min(double x, double y)
{
//比较两个数的大小，返回小的数
if(x < y) return x;
else return y;
}
bool Intersect(TSegment L1, TSegment L2)
{
//线段l1与l2相交而且不在端点上时，返回true
//判断线段是否相交
//1.快速排斥试验判断以两条线段为对角线的两个矩形是否相交
TPoint s1 = L1.p1;
TPoint e1 = L1.p2;
TPoint s2 = L2.p1;
TPoint e2 = L2.p2;
//2.跨立试验
if(
(max(s1.x, e1.x) > min(s2.x, e2.x)) &&
(max(s2.x, e2.x) > min(s1.x, e1.x)) &&
(max(s1.y, e1.y) > min(s2.y, e2.y)) &&
(max(s2.y, e2.y) > min(s1.y, e1.y)) &&
(multi(s2, e1, s1) * multi(e1, e2, s1) > 0) &&
(multi(s1, e2, s2) * multi(e2, e1, s2) > 0)
) return true;
return false;
}
bool Online(TSegment L, TPoint p)
{
//p在L上(不在端点)时返回true
//1.在L所在的直线上
//2.在L为对角线的矩形中
double dx, dy, dx1, dy1;
dx = L.p2.x - L.p1.x;
dy = L.p2.y - L.p1.y;
dx1 = p.x - L.p1.x;
dy1 = p.y - L.p1.y;
if(dx * dy1 - dy * dx1 != 0) return false; //叉积
if(dx1 * (dx1 - dx) < 0 || dy1 * (dy1 - dy) < 0) return true;
return false;
}
bool same1(TSegment L, TPoint p1, TPoint p2)
{
//判断p1, p2是否在L的同侧
if(multi(p1, L.p2, L.p1) * multi(L.p2, p2, L.p1)< 0) return true;
return false;
}
bool Inside(TPoint q, TPolygon polygon)
{
int c, i;
TSegment L1, L2;
c = 0;
L1.p1 = q;
L1.p2 = q;
L1.p2.x = INF;
/**//*
（1）相交
1.p[i]和p[i+1]在L的两侧
2.p[i]和p[i+2]在L的同侧
3.p[u]和p[i+3]在L的同侧或异侧
*/
for(i = 0;i <= polygon.n - 1;i++){
L2.p1 = polygon.point[i];
L2.p2 = polygon.point[(i + 1) % polygon.n];
if(Intersect(L1, L2)){
c++;
continue;
}
if(!Online(L1, polygon.point[(i + 1) % polygon.n])) continue;
if(!Online(L1, polygon.point[(i + 2) % polygon.n]) &&
!same1(L1, polygon.point[i], polygon.point[(i + 2) % polygon.n])){
c++;
continue;
}
if(Online(L1, polygon.point[(i + 2) % polygon.n]) &&
!same1(L1, polygon.point[i], polygon.point[(i + 3) % polygon.n]))
c++;
}
if(c % 2 == 0) return false;
else return true;
}
int main()
{
int i, test, k;
int primp, primq;
TPoint p;
p.x = 0;
p.y = 0;
test = 1;
TPolygon polygon;
while(scanf("%d", &polygon.n) != EOF && polygon.n){
printf("Pilot %d\n", test++);
for(i = 0;i <= polygon.n - 1;i++){
scanf("%lf%lf", &polygon.point[i].x, &polygon.point[i].y);
}
scanf("%d%d", &primp, &primq);
if(Inside(p, polygon)){
printf("The pilot is in danger!\n");
k = (primp - 1) * (primq - 1) / 2;
printf("The secret number is %d.\n", k);
}
else printf("The pilot is safe.\n");
printf("\n");
}
return 0;
}posted on 2011-02-06 22:55
小阮 阅读(321)
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