https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2640

http://www.spoj.com/problems/SPOINTS/en/

http://poj.org/problem?id=3805

http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1298

要想有一条直线分两个凸包,两个凸包不相交,不相切是必要的

在没有模板的情况下,我的代码,过了poj,uva,和spoj的,但是过不了aoj的,和正确代码对拍所发现的情况不太符合事实

方法是:

1.判断每个点是否在凸包里面或者边上

2.判断两凸包每两条线段是否相交

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const double eps=1e-8;
const int maxn=205;
int dcmp(double d){
if(fabs(d)<eps)return 0;
return d>0?1:-1;
}
struct pnt{
double x,y;
pnt():x(0),y(0){}
pnt(double tx,double ty):x(tx),y(ty){}
pnt operator -(pnt p2){
pnt newp(x-p2.x,y-p2.y);
return newp;
}
pnt operator +(pnt p2){
pnt newp(x+p2.x,y+p2.y);
return newp;
}
pnt operator *(double d){
pnt newp(x*d,y*d);
return newp;
}
pnt operator /(double d){
pnt newp(x/d,y/d);
return newp;
}
double dis(pnt p2){
return sqrt((x-p2.x)*(x-p2.x)+(y-p2.y)*(y-p2.y));
}
bool operator ==(pnt p2){
if(dcmp(x-p2.x)==0&&dcmp(y-p2.y)==0)return true;
return false;
}
};
double cross(pnt p1,pnt p2){
return p1.x*p2.y-p1.y*p2.x;
}
bool cmpx(pnt p1,pnt p2){
if(p1.x!=p2.x)return p1.x<p2.x;
return p1.y<p2.y;
}
pnt base;
bool cmp(pnt p1,pnt p2){
return cross(p1- base,p2-base)<0;
} int isPointInConvexPolygon(pnt p1,pnt * p,int n){
for(int i=0;i<n;i++){
pnt A=pnt(p[(i+1)%n].x-p[i].x,p[(i+1)%n].y-p[i].y);
pnt B=pnt(p1.x-p[i].x,p1.y-p[i].y);
int fl=dcmp(cross(A,B));
if(fl<0)return 0;
if(fl==0){
int maxx=max(p[(i+1)%n].x,p[i].x);
int minx=min(p[(i+1)%n].x,p[i].x);
int maxy=max(p[(i+1)%n].y,p[i].y);
int miny=min(p[(i+1)%n].y,p[i].y);
if(minx<=p1.x&&maxx>=p1.x&&miny<=p1.y&&maxy>=p1.y)return -1;//on the edge
else return 0;
}
}
return 1;
}
int graham(pnt * p,pnt * h,int n){ int m=0;
for(int i=0;i<n;i++){//计算上凸包
while(m>1&&cross((h[m-1]-h[m-2]),(p[i]-h[m-2]))<=0){m--;}
h[m++]=p[i];
}
int tm=m;
for(int i=n-2;i>=0;i--){//计算下凸包
while(m>tm&&cross((h[m-1]-h[m-2]),(p[i]-h[m-2]))<=0){m--;}
h[m++]=p[i];
}
if(n>1)m--;
return m;
} bool between(pnt p1,pnt p2,pnt p){
return (p.x<=max(p1.x,p2.x)&&p.x>=min(p1.x,p2.x))
&& (p.y<=max(p1.y,p2.y)&&p.y>=min(p1.y,p2.y));
}
bool isInsert(pnt p11,pnt p12,pnt p21,pnt p22){
pnt v=p22-p21;
pnt w=p12-p11;
pnt u=p21-p11;
if(cross(v,w)==0){
if(cross(v,u)!=0)return false;
if(between(p11,p12,p21))return true;
if(between(p11,p12,p22))return true;
return false;
}
double t=cross(w,u)/cross(v,w);
double t2=cross(v,u)/cross(v,w);
if(t2>1||t2<0)return false;
if(t>1||t<0)return false;
return true;
} pnt bp[maxn],bh[maxn],wp[maxn],wh[maxn];
int bn,wn,btop,wtop; int main(){
while(scanf("%d%d",&bn,&wn)==2&&(bn||wn)){
for(int i=0;i<bn;i++){
scanf("%lf%lf",&bp[i].x,&bp[i].y);
}
for(int i=0;i<wn;i++){
scanf("%lf%lf",&wp[i].x,&wp[i].y);
}
bool fl=true; if(bn>1){
sort(bp,bp+bn,cmpx);
base =bp[0];
sort(bp+1,bp+bn,cmp);
btop=graham(bp,bh,bn); if(fl)for(int i=0;i<wn;i++){
if(isPointInConvexPolygon(wp[i],bh,btop)){
fl=false;
break;
}
}
} if(wn>1){
sort(wp,wp+wn,cmpx);
base =wp[0];
sort(wp+1,wp+wn,cmp);
wtop=graham(wp,wh,wn); if(fl){
for(int i=0;i<bn;i++){
if(isPointInConvexPolygon(bp[i],wh,wtop)){
fl=false;
break;
}
}
}
} if(fl&&bn>1&&wn>1){
for(int i=0;i<btop;i++){
for(int j=0;j<wtop;j++){
if(isInsert(bh[i],bh[(i+1)%btop],wh[j],wh[(j+1)%wtop])){
fl=false;
break;
}
}
}
}
if(wn==1&&bn==1&&bp[0]==wp[0]){fl=false;} if(fl)puts("YES");
else puts("NO");
}
return 0;
}

  这里的代码都可以过:http://pelkira.hatenablog.jp/

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<vector>
#include<queue>
#include<functional>
#include<cstring>
#include<cstdlib>
#include<complex>
using namespace std;
typedef long long ll;
typedef complex < double > Point; typedef vector < Point > Polygon;
namespace std {
bool operator < (Point a,Point b) {
return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
}
}
const double EPS = 1e-8;
const double INF = 1e12; struct Line : Polygon{
Line(){};
Line(Point p,Point q){ push_back(p);push_back(q); }
};
double cross(Point p,Point q){
return imag(conj(p)*q);
}
double dot(Point p,Point q){
return real(conj(p)*q);
} int ccw(Point a,Point b,Point c){
b-=a,c-=a;
if(cross(b,c)>0)return 1; //反時計回り
if(cross(b,c)<0)return -1; //時計回り
if(dot(b,c)<0)return 2; //直線上に c - a - b
if(norm(b)<norm(c))return -2; // 直線上に a - b - c
return 0; // 直線上に a - c - b
} bool intersectSS(Line s, Line t) {
return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}
bool intersectSP(Line s, Point p) {
return abs(s[0]-p)+abs(s[1]-p)-abs(s[1]-s[0]) < EPS; // triangle inequality
}
Polygon ConvexHull(Polygon ps) {//凸包
int n = ps.size(), k = 0;
if(n == 1)return ps;
sort(ps.begin(), ps.end());
Polygon ch(2*n);
for (int i = 0; i < n; ch[k++] = ps[i++]) // lower-hull
while (k >= 2 && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;
for (int i = n-2, t = k+1; i >= 0; ch[k++] = ps[i--]) // upper-hull
while (k >= t && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;
ch.resize(k-1);
return ch;
} // Point - Vertex
enum{OUT, ON, IN};
int IsInnerPointVertex(Polygon ps,Point a){
bool flg = false;
for(int i = 0;i < ps.size();i++){
Point p = ps[i] - a,q = ps[(i+1)%ps.size()] - a;
if(imag(p)>imag(q))swap(p,q);
if(imag(p)<=0&&0<imag(q)){
if(cross(p,q) < 0)flg = !flg;
}
if(cross(p,q) == 0 && dot(p,q) <= 0)return ON;
}
return flg ? IN : OUT;
}
int main(){
int n,m;
while(cin>>n>>m,n){
Polygon G,H;
for(int i = 0;i < n;i++){
double x,y;
cin>>x>>y;
G.push_back(Point(x,y));
}
for(int i = 0;i < m;i++){
double x,y;
cin>>x>>y;
H.push_back(Point(x,y));
}
G = ConvexHull(G);
H = ConvexHull(H);
int g = G.size(),h = H.size();
//cout<<g<<" "<<h<<endl;
if(g > h)swap(G,H);
g = G.size(),h = H.size();
bool flg = true;
for(int i = 0;i < g;i++){
if(IsInnerPointVertex(H,G[i]) != OUT){
flg = false;
break;
}
}
for(int i = 0;i < h;i++){
if(IsInnerPointVertex(G,H[i]) != OUT){
flg = false;
break;
}
} if(h == 1)flg = true;
else if(g == 1 && h == 2){
flg = !(intersectSP(Line(H[0],H[1]),G[0]));
}
else if(g == 2 && h == 2){
flg = !(intersectSS(Line(G[0],G[1]),Line(H[0],H[1])));
}
else if(g == 2){
flg = true;
for(int i = 0;i < h;i++){
int j = (i+1)%h;
if(intersectSS(Line(G[0],G[1]),Line(H[i],H[j])))flg = false;
}
}
if(flg)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}

  

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