题意：
顺时针给你n个点，然后m个询问，问你给出的点是否在n个点构成的多边形的内部。
题解：
直接上模板。
1 #include<bits/stdc++.h> 2 #define F(i,a,b) for(int i=a;i<=b;++i) 3 #define MAXN 1007 4 #define offset 10000 5 #define eps 1e-10 6 //浮点数判 0 7 #define zero(x) (((x)>0?(x):-(x))<eps) 8 //浮点数判断符 9 #define _sign(x) ((x)>eps?1:((x)<-eps?2:0)) 10 //定义点 11 struct point 12 { 13 double x,y; 14 }pt[MAXN],tmp; 15 double xmult(point p1,point p2,point p0) 16 { 17 return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); 18 } 19 int inside_polygon(point q,int n,point* p,int on_edge=2) 20 { 21 point q2; 22 int i=0,count; 23 while (i<n) 24 for (count=i=0,q2.x=rand()+offset,q2.y=rand()+offset;i<n;i++) 25 { 26 if (zero(xmult(q,p[i],p[(i+1)%n]))&&(p[i].x-q.x)*(p[(i+1)%n].x-q.x)<eps&&(p[i].y-q.y)*(p[(i+1)%n].y-q.y)<eps) 27 return on_edge; 28 else if (zero(xmult(q,q2,p[i])))break; 29 else if (xmult(q,p[i],q2)*xmult(q,p[(i+1)%n],q2)<-eps&& 30 xmult(p[i],q,p[(i+1)%n])*xmult(p[i],q2,p[(i+1)%n])<-eps) 31 count++; 32 } 33 return count&1; 34 } 35 int n,m; 36 int main() 37 { 38 while(~scanf("%d",&n)) 39 { 40 F(i,0,n-1)scanf("%lf%lf",&pt[i].x,&pt[i].y); 41 scanf("%d",&m); 42 F(i,1,m) 43 { 44 scanf("%lf%lf",&tmp.x,&tmp.y); 45 puts(inside_polygon(tmp,n,pt)?"Yes":"No"); 46 } 47 } 48 return 0; 49 }