UVA 12931 Common Area

题意：给出两个多边形，判断是否有公共面积。

做法：很显然若不重合，必然有一个多边形有边穿过或就在另一个多边形里面。问题就很简单了。

#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;const double eps = 1e-8;const double pi = acos(-1.0);int sgn(double x){return fabs(x)<eps?0:x<0?-1:1;}struct point{double x,y;point(){}point(double x,double y){this->x=x;this->y=y;}point operator +(point a){return point(x+a.x,y+a.y);}point operator -(point a){return point(x-a.x,y-a.y);}double operator *(point a){return x*a.x + y*a.y;}double operator /(point a){return x*a.y-y*a.x;}bool operator <(point a)const{return x!=a.x?x<a.x:y<a.y;}};struct line{point s,e;line(){}line(point s,point e){this->s=s;this->e=e;}//????????//?????0??????,?1????,?0????,?2???//???????2?,??????pair<int,point> operator *(line a){point res = s;if(sgn((s-e)/(a.s-a.e))==0){if(sgn((s-a.e)/(a.s-a.e))==0)return make_pair(0,res);//??return make_pair(1,res);//??}double t=(s-a.s)/(a.s-a.e)/((s-e)/(a.s-a.e));res.x+=(e.x-s.x)*t;res.y+=(e.y-s.y)*t;return make_pair(2,res);}};//a?b??c point transxy(point a,point b,double c){point t1=a-b,t2=point(sin(c),cos(c));return b+point(t1/t2,t1*t2);}//?????double dis(point a,point b){point t=a-b;return sqrt(t*t);}//??????bool inter(line l1,line l2){returnmax(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&sgn((l2.s-l1.e)/(l1.s-l1.e))*sgn((l2.e-l1.e)/(l1.s-l1.e))<=0&&sgn((l1.s-l2.e)/(l2.s-l2.e))*sgn((l1.e-l2.e)/(l2.s-l2.e))<=0;}//????l1???l2????bool seg_inter_line(line l1,line l2) {return sgn((l2.s-l1.e)/(l1.s-l1.e))*sgn((l2.e-l1.e)/(l1.s-l1.e))<=0;}//??????????res,?????????point point_to_line(point p,line l){point res;double t=(p-l.s)*(l.e-l.s)/((l.e-l.s)*(l.e-l.s));res.x=l.s.x+(l.e.x-l.s.x)*t;res.y=l.s.y+(l.e.y-l.s.y)*t;return res;}//??????????????????point point_to_seg(point p,line l){point res;double t =(p-l.s)*(l.e-l.s)/((l.e-l.s)*(l.e-l.s));if(t>=0&&t<=1){res.x=l.s.x+(l.e.x - l.s.x)*t;res.y=l.s.y+(l.e.y - l.s.y)*t;}elseres=dis(p,l.s)<dis(p,l.e)?l.s:l.e;return res;}//???????double calc_area(point p[],int n){double res = 0;for(int i = 0;i < n;i++)res += (p[i]/p[(i+1)%n])/2;return fabs(res);}//???????bool onseg(point p,line l){returnsgn((l.s-p)/(l.e-p))==0&&sgn((p.x-l.s.x)*(p.x-l.e.x))<=0&&sgn((p.y-l.s.y)*(p.y-l.e.y))<=0;}//?????????//???????,????????(??????????<0??>0)int inconvexpoly(point a,point p[],int n){for(int i = 0;i < n;i++){if(sgn((p[i]-a)/(p[(i+1)%n]-a))<0)return 1;//???? if(onseg(a,line(p[i],p[(i+1)%n])))return 0;//?????}return -1;//????}//??????????int point_in_poly(point a,point p[],int n){int wn=0;for(int i=0;i<n;i++){if(onseg(a,line(p[i],p[(i+1)%n])))return 0;//????? int t1=sgn((p[(i+1)%n]-p[i])/point(a-p[i]));int t2=sgn(p[i].y-a.y),t3=sgn(p[(i+1)%n].y-a.y);if(t1>0&&t2<=0&&t3>0) wn++;if(t1<0&&t3<=0&&t2>0) wn--;}return wn!=0?-1:1;//-1???,1??? }//??????//?????//??????????????????//????1~n-1bool isconvex(point p[],int n){bool s[3];memset(s,false,sizeof(s));for(int i=0;i<n;i++){s[sgn((p[(i+1)%n]-p[i])/(p[(i+2)%n]-p[i]))+1]=1;if(s[0] && s[2])return 0;}return 1;}//??int covexhull(point p[],int n,point ch[]){sort(p,p+n);int m=0;for(int i=0;i<n;i++){while(m>1&&sgn((ch[m-1]-ch[m-2])/(p[i]-ch[m-2]))<=0)m--;ch[m++]=p[i];}int k=m;for(int i=n-2;i>-1;i--){while(m>k&&sgn((ch[m-1]-ch[m-2])/(p[i]-ch[m-2]))<=0)m--;ch[m++]=p[i];}if(n>1)m--;return m;}point bx[2][110];int main(){int n,m,cs=0;while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++)scanf("%lf%lf",&bx[0][i].x,&bx[0][i].y);scanf("%d",&m);for(int i=0;i<m;i++)scanf("%lf%lf",&bx[1][i].x,&bx[1][i].y);bool flag=0;for(int i=0;!flag&&i<n;i++){line a=line(bx[0][i],bx[0][(i+1)%n]);vector<point>bb;bb.push_back(bx[0][i]);bb.push_back(bx[0][(i+1)%n]);for(int j=0;j<m;j++){line b=line(bx[1][j],bx[1][(j+1)%m]);if(!inter(a,b))continue; pair<int,point> res=a*b;if(res.first==2)bb.push_back(res.second);}sort(bb.begin(),bb.end());int len=bb.size();for(int j=0;!flag&&j<len;j++){point t=bb[j]+bb[(j+1)%len];t.x/=2;t.y/=2;if(point_in_poly(bb[j],bx[1],m)<0||point_in_poly(t,bx[1],m)<0)flag=1;}}for(int i=0;!flag&&i<m;i++){line a=line(bx[1][i],bx[1][(i+1)%m]);vector<point>bb;bb.push_back(bx[1][i]);bb.push_back(bx[1][(i+1)%m]);for(int j=0;j<n;j++){line b=line(bx[0][j],bx[0][(j+1)%n]);if(!inter(a,b))continue; pair<int,point> res=a*b;if(res.first==2)bb.push_back(res.second);}sort(bb.begin(),bb.end());int len=bb.size();for(int j=0;!flag&&j<len;j++){point t=bb[j]+bb[(j+1)%len];t.x/=2;t.y/=2;if(point_in_poly(bb[j],bx[0],n)<0||point_in_poly(t,bx[0],n)<0)flag=1;}}if(!flag){bool flag1=0;for(int i=0;!flag1&&i<n;i++)if(point_in_poly(bx[0][i],bx[1],m)==1)flag1=1;if(!flag1)flag=1;}printf("Case %d: %s\n",++cs,flag?"Yes":"No");}}

Given two simple polygons, your task is to determine whether they have a non-empty common area.
Note that the two rectangles in gure (a) share a segment, but no common area at all.
By \simple polygon", we mean the polygons will not be self-intersecting or self-touching, and will
not have duplicated vertices or adjacent collinear segments.
Note: be sure to test your program with many special cases.
Input
There will be at most 100 test cases. Each test case contains two lines, one for each polygon. Each
polygon begins with an integer n (3  n  100), the number of vertices, then n pairs of integers (x; y)
(