USACO JAN14 奶牛冰壶运动 凸包+判定
2024-10-13 11:50:41
满足条件的一定是在凸包内的,直接判断
恬不知耻的加了特判,2333
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 50050
using namespace std;
int n,ss[N],top,topa,topb,ans1,ans2;
bool bo=0;
struct point{
double x,y;
}a[N],b[N],o,p[N];
double operator * (point a,point b){return a.x*b.y-b.x*a.y;}
point operator - (point a,point b){return (point){a.x-b.x,a.y-b.y};}
double dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
bool operator < (point a,point b){
double s=(a-o)*(b-o);
if(s==0) return dis(a,o)<dis(b,o);
return s>0;
}
void graham(point *gy){
ss[++top]=1;
for(int i=2;i<=n;i++){
while(top>1&&(gy[i]-gy[ss[top-1]])*(gy[ss[top]]-gy[ss[top-1]])>0) top--;
ss[++top]=i;
}
}
bool checka(point b){
int l=1,r=topa,mid;
while(l<=r){
mid=(l+r)>>1;
double a1=(a[ss[mid]]-a[ss[1]])*(b-a[ss[1]]);
double a2=(a[ss[mid+1]]-a[ss[1]])*(b-a[ss[1]]);
if(a1>=0&&a2<=0){
if((a[ss[mid+1]]-a[ss[mid]])*(b-a[ss[mid]])>=0) return 1;
return 0;
}
else{
if(a1<0) r=mid-1;
else l=mid+1;
}
}
return 0;
}
bool checkb(point a){
int l=topa+1,r=topb,mid;
while(l<=r){
mid=(l+r)>>1;
double a1=(b[ss[mid]]-b[ss[topa+1]])*(a-b[ss[topa+1]]);
double a2=(b[ss[mid+1]]-b[ss[topa+1]])*(a-b[ss[topa+1]]);
if(a1>=0&&a2<=0){
if((b[ss[mid+1]]-b[ss[mid]])*(a-b[ss[mid]])>=0) return 1;
return 0;
}
else{
if(a1<0) r=mid-1;
else l=mid+1;
}
}
return 0;
}
int main()
{
freopen("curling.in","r",stdin);
freopen("curling.out","w",stdout);
scanf("%d",&n);
int yx=1;
for(int i=1;i<=n;i++){
scanf("%lf%lf",&a[i].x,&a[i].y);
if((a[i].x<a[yx].x)||(a[i].x==a[yx].x&&a[i].y<a[yx].y))yx=i;
if(a[i].x!=0) bo=1;
}
o=a[yx]; swap(a[1],a[yx]);
sort(a+2,a+n+1);
//for(int i=1;i<=n;i++)p[i]=a[i];
graham(a); topa=top;
yx=1;
for(int i=1;i<=n;i++){
scanf("%lf%lf",&b[i].x,&b[i].y);
if((b[i].x<b[yx].x)||(b[i].x==b[yx].x&&b[i].y<b[yx].y))yx=i;
if(b[i].x!=0) bo=1;
}
if(!bo){
printf("%d %d\n",n-1,n-1);
return 0;
}
o=b[yx]; swap(b[1],b[yx]);
sort(b+2,b+n+1);
graham(b); topb=top;
for(int i=1;i<=n;i++)
if(checka(b[i]))
ans1++;
for(int i=1;i<=n;i++)
if(checkb(a[i]))
ans2++;
printf("%d %d\n",ans1,ans2);
return 0;
}
最新文章
- 同步降压DC-DC转换IC——XC9264
- php性能剖析的几款软件
- 一款全兼容的播放器 videojs
- C#的数组
- js-JavaScript高级程序设计学习笔记6
- Power Strings 分类: POJ 串 2015-07-31 19:05 8人阅读 评论(0) 收藏
- delphi快捷键
- Android Service实时向Activity传递数据
- DS18B20 for STM32 源代码 【worldsing笔记】
- 关于js跨域
- 正在搞用web.py做的通讯录
- resultMap之collection聚集
- iOS中的base64加密
- IOS CrackMe 破解学习
- Android多线程下安全访问数据库
- 视频云SDK iOS持续集成项目实践
- 大区间素数筛选(POJ 2689)
- 【转】globk中的卫星轨道约束
- 新版MATERIAL DESIGN 官方动效指南(二)
- Python的易错点