【kuangbin专题】计算几何_凸包

1.poj1113 Wall

题目：http://poj.org/problem?id=1113

题意：用一条线把若干个点包起来，并且线距离任何一个点的距离都不小于r。求这条线的最小距离是多少？

分析：这道题的答案是凸包周长加上一个圆周长，即包围凸包的一个圆角多边形，但是没弄明白那些圆角加起来为什么恰好是一个圆。每个圆角是以凸包对应的顶点为圆心，给定的L为半径，与相邻两条边的切点之间的一段圆弧。每个圆弧的两条半径夹角与对应的凸包的内角互补。假设凸包有n条边，则所有圆弧角之和为180°*n-180°*（n-2）=360°。故，围墙周长为=n条平行于凸包的线段+n条圆弧的长度=凸包周长+围墙离城堡距离L为半径的圆周长。

 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const int maxn=;
 const double eps=1e-;
 const double pi=acos(-1.0);
 inline double sqr(double x){return x*x;}
 int sgn(double x){
     if (fabs(x)<eps) return ;
     if (x<) return -;
     return ;
 }
 struct point{
     double x,y;
     point(){}
     point(double _x,double _y):x(_x),y(_y){}
     //判断两点相同
     bool operator ==(const point &b) const{
         return sgn(x-b.x)== && sgn(y-b.y)==;
     }
     //
     point operator -(const point &b) const{
         return point(x-b.x,y-b.y);
     }
     //叉积
     double operator ^(const point &b) const{
         return x*b.y-y*b.x;
     }
     //点积
     double operator *(const point &b) const{
         return x*b.x+y*b.y;
     }
     //重载小于号   （求最左下角的点
     bool operator <(const point &b)const{
         return sgn(x-b.x)== ? sgn(y-b.y)< : x<b.x;
     }
 };
 struct line{
     point s,e;
     line(){}
     line(point _s,point _e):s(_s),e(_e){}
 };
 double dis(point a,point b){
     return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
 }
 struct polygon{
     int n;
     point p[maxn];
     line l[maxn];
     void add(point q){
         p[n++]=q;
     }
     void getline(){
         for (int i=;i<n;i++)
             l[i]=line(p[i],p[(i+)%n]);
     }
     struct cmp{
         point p;
         cmp(const point &p0):p(p0){}
         bool operator ()(const point &aa,const point &bb){
             point a=aa,b=bb;
             int d=sgn((a-p)^(b-p));
             if (d==){
                 return sgn(dis(a,p)-dis(b,p))<;
             }
             return d>;
         }
     };
     //极角排序  先找到左下角的点
     //重载好point的'<'
     void norm(){
         point mi=p[];
         for (int i=;i<n;i++) mi=min(mi,p[i]);
         sort(p,p+n,cmp(mi));
     }
     //得到凸包，点编号为0--n-1
     void Graham(polygon &convex){
         norm();
         int &top=convex.n;
         top=;
         if (n==){
             top=; convex.p[]=p[]; return ;
         }
         if (n==){
             top=; convex.p[]=p[]; convex.p[]=p[];
             if (convex.p[]==convex.p[]) top--;
             return ;
         }
         convex.p[]=p[]; convex.p[]=p[]; top=;
         for (int i=;i<n;i++){
             while (top> && sgn((convex.p[top-]-convex.p[top-])^(p[i]-convex.p[top-]))<=) top--;
             convex.p[top++]=p[i];
         }
         if (convex.n== && (convex.p[]==convex.p[])) convex.n--;
     }
 };
 polygon C;
 int main(){
     int n,L; cin >> n >> L;
     double x,y;
     for (int i=;i<n;i++){
         cin >> x >> y;
         C.add(point(x,y));
     }
     polygon ans;
     C.Graham(ans);
     ans.getline();
     double res=;
     for (int i=;i<ans.n;i++) res+=dis(ans.l[i].s,ans.l[i].e);
     res+=*pi*L;
     printf("%d\n",(int)(res+0.5));
     return ;
 }

poj1113

（为了套板子，写的很繁琐）

2.poj2007 Scrambled Polygon

题目：http://poj.org/problem?id=2007

题意：求一个凸多边形的凸包。要求起始点为输入的第一个点。

分析：rt。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 using namespace std;
 const double eps=1e-;
 const int maxn=;
 inline double sqr(int x){return x*x*1.0;}
 int sgn(double x){
     if (fabs(x)<eps) return ;
     if (x<) return -;
     return ;
 }
 struct point{
     int x,y;
     point(){}
     point(int _x,int _y):x(_x),y(_y){}
     bool operator ==(const point &b)const{
         return (x==b.x && y==b.y);
     }
     point operator -(const point &b)const{
         return point(x-b.x,y-b.y);
     }
     int operator ^(const point &b)const{
         return x*b.y-y*b.x;
     }
     int operator *(const point &b)const{
         return x*b.x+y*b.y;
     }
     //重载小于号   （求最左下角的点
     bool operator <(const point &b)const{
         return sgn((x-b.x)*1.0)== ? sgn((y-b.y)*1.0)< : x<b.x;
     }
 };
 double dis(point a,point b){
     return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
 }
 struct polygon{
     int n;
     point p[maxn];
     void add(point q){
         p[n++]=q;
     }
     struct cmp{
         point p;
         cmp(const point &p0):p(p0){}
         bool operator ()(const point &aa,const point &bb){
             point a=aa,b=bb;
             int k=(a-p)^(b-p);int d;
             if (k==) d=;else if (k<) d=-; else d=;
             if (d==){
                 return sgn(dis(a,p)-dis(b,p))<;
             }
             return d>;
         }
     };
     //极角排序  先找到左下角的点
     //重载好point的'<'
     void norm(){
         point mi=p[];
         for (int i=;i<n;i++) mi=min(mi,p[i]);
         sort(p,p+n,cmp(mi));
     }
     //得到凸包，点编号为0--n-1
     void Graham(polygon &convex){
         norm();
         int &top=convex.n;
         top=;
         if (n==){
             top=; convex.p[]=p[]; return ;
         }
         if (n==){
             top=; convex.p[]=p[]; convex.p[]=p[];
             if (convex.p[]==convex.p[]) top--;
             return ;
         }
         convex.p[]=p[]; convex.p[]=p[]; top=;
         for (int i=;i<n;i++){
             while (top> && sgn(((convex.p[top-]-convex.p[top-])^(p[i]-convex.p[top-]))*1.0)<=) top--;
             convex.p[top++]=p[i];
         }
         if (convex.n== && (convex.p[]==convex.p[])) convex.n--;
     }
 };
 polygon C;
 int main(){
     int x,y; point p;
     C.n=;
     cin >> x >> y; p=point(x,y); C.add(p);
     while (cin >> x >> y) C.add(point(x,y));
     polygon ans;
     C.Graham(ans); int k;
     for (int i=;i<ans.n;i++) if (ans.p[i]==p){k=i;break;}
     for (int i=;i<ans.n;i++){
         printf("(%d,%d)\n",ans.p[(k+i)%ans.n].x,ans.p[(k+i)%ans.n].y);
     }
     return ;
 }

poj2007

3.poj1228

　　题目：http://poj.org/problem?id=1228

题意：输入一个凸包上的点（没有凸包内部的点，要么是凸包顶点，要么是凸包边上的点），判断这个凸包是否稳定。所谓稳定就是判断能不能在原有凸包上加点。

分析：问凸包是否稳定。

 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const double eps=1e-;
 const int maxn=;
 int sgn(double x){
     if (fabs(x)<eps) return ;
     if (x<) return -;
     return ;
 }
 struct point{
     double x,y;
     point(){}
     point(double _x,double _y):x(_x),y(_y){}
     point operator -(const point &b)const{
         return point(x-b.x,y-b.y);
     }
     double operator ^(const point &b)const{
         return x*b.y-y*b.x;
     }
 }p[maxn],pp[maxn];
 bool cmp(point a,point b) {return a.x < b.x || (a.x == b.x && a.y < b.y);}
 int convexhull(point p[],int n,point pp[]){
     sort(p,p+n,cmp);
     int m=;
     for (int i=;i<n;i++){
         while (m> && ((pp[m-]-pp[m-])^(p[i]-pp[m-]))<) m--;
         pp[m++]=p[i];
     }
     int k=m;
     for (int i=n-;i>=;i--){
         while (m>k && ((pp[m-]-pp[m-])^(p[i]-pp[m-]))<) m--;
         pp[m++]=p[i];
     }
     return m-;
 }
 bool check(point p[],int n){
     for (int i=;i<n-;i++){
         if (((p[i-]-p[i])^(p[i+]-p[i]))!= &&
             ((p[i]-p[i+])^(p[i+]-p[i+]))!=) return false;  //保证至少有三个点在同一条边上
     }
     return true;
 }
 int main(){
     int t,n; double x,y; cin >> t;
     while (t--){
         cin >> n;
         for (int i=;i<n;i++) cin >> p[i].x >> p[i].y;
         if (n<){cout << "NO\n";continue;}
         int cnt=convexhull(p,n,pp);
         if (check(pp,cnt)) cout << "YES\n"; else cout << "NO\n";
     }
     return ;
 }

poj1228

4

5.poj3348 Cows

题目：http://poj.org/problem?id=3348

题意：给出一些点圈出一个最大面积每50平方养一头牛问最多能养多少牛。

分析：凸包+多边形面积。

 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const int maxn=;
 inline int sqr(int x){return x*x;}
 struct point{
     int x,y;
     point(){}
     point(int _x,int _y):x(_x),y(_y){}
     //判断两点相同
     bool operator ==(const point &b) const{
         return x-b.x== && y-b.y==;
     }
     //
     point operator -(const point &b) const{
         return point(x-b.x,y-b.y);
     }
     //叉积
     int operator ^(const point &b) const{
         return x*b.y-y*b.x;
     }
     //点积
     int operator *(const point &b) const{
         return x*b.x+y*b.y;
     }
     //重载小于号   （求最左下角的点
     bool operator <(const point &b)const{
         return x-b.x==?y-b.y<:x<b.x;
     }
 };
 struct line{
     point s,e;
     line(){}
     line(point _s,point _e):s(_s),e(_e){}
 };
 int dis(point a,point b){
     return (int)sqrt(1.0*sqr(a.x-b.x)+1.0*sqr(a.y-b.y));
 }
 struct polygon{
     int n;
     point p[maxn];
     void add(point q){p[n++]=q;}
     struct cmp{
         point p;
         cmp(const point &p0):p(p0){}
         bool operator ()(const point &aa,const point &bb){
             point a=aa,b=bb;
             int k=(a-p)^(b-p);
             if (k==){
                 return dis(a,p)-dis(b,p)<;
             }
             return k>;
         }
     };
     //极角排序  先找到左下角的点
     //重载好point的'<'
     void norm(){
         point mi=p[];
         for (int i=;i<n;i++) mi=min(mi,p[i]);
         sort(p,p+n,cmp(mi));
     }
     //得到凸包，点编号为0--n-1
     void Graham(polygon &convex){
         norm();
         int &top=convex.n;
         top=;
         if (n==){
             top=; convex.p[]=p[]; return ;
         }
         if (n==){
             top=; convex.p[]=p[]; convex.p[]=p[];
             if (convex.p[]==convex.p[]) top--;
             return ;
         }
         convex.p[]=p[]; convex.p[]=p[]; top=;
         for (int i=;i<n;i++){
             while (top> && ((convex.p[top-]-convex.p[top-])^(p[i]-convex.p[top-]))<=) top--;
             convex.p[top++]=p[i];
         }
         if (convex.n== && (convex.p[]==convex.p[])) convex.n--;
     }
     int getarea(){
         int sum=;
         for (int i=;i<n;i++) sum+=(p[i]^(p[(i+)%n]));
         return sum/;
     }
 };
 polygon C;
 int main(){
     int n,x,y; cin >> n; C.n=;
     for (int i=;i<n;i++){
         cin >> x >> y;
         C.add(point(x,y));
     }
     polygon ans; C.Graham(ans);
     cout << ans.getarea()/ << endl;
     return ;
 }

poj3348

6.poj1259/hdoj6219

题意：给出n个点，求出最大面积的凸包且凸包内不包含原点集中的点。

分析：rt。求最大面积空凸包。

 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const double eps=1e-;
 const int maxn=;
 inline double sqr(double x){return x*x;}
 int sgn(double x){
     if (fabs(x)<eps) return ;
     if (x<) return -;
     return ;
 }
 struct point{
     double x,y;
     point(){}
     point(double _x,double _y):x(_x),y(_y){}
     //判断两点相同
     bool operator ==(const point &b) const{
         return sgn(x-b.x)== && sgn(y-b.y)==;
     }
     //
     point operator -(const point &b) const{
         return point(x-b.x,y-b.y);
     }
     //叉积
     double operator ^(const point &b) const{
         return x*b.y-y*b.x;
     }
     //点积
     double operator *(const point &b) const{
         return x*b.x+y*b.y;
     }
     //重载小于号   （求最左下角的点
     bool operator <(const point &b)const{
         return sgn(x-b.x)== ? sgn(y-b.y)< : x<b.x;
     }
 }p[maxn],pp[maxn];
 point tmp;
 double dp[maxn][maxn];
 double empty_convex(point p[],int n,point o){
     double ans=;
     for (int i=;i<n;i++) for (int j=;j<n;j++) dp[i][j]=;
     for (int i=;i<n;i++){
         int j=i-;
         while (j>= && ((p[i]-o)^(p[j]-o))==) j--;
         bool flag=(j==i-);
         while (j>=){
             int k=j-;
             while (k>= && ((p[i]-p[k])^(p[j]-p[k]))>) k--;
             double area=fabs((p[i]-o)^(p[j]-o))/2.0;
             if (k>=) area+=dp[j][k];
             if (flag) dp[i][j]=area;
             ans=max(ans,area);
             j=k;
         }
         if (flag){
             for (int j=;j<i;j++) dp[i][j]=max(dp[i][j],dp[i][j-]);
         }
     }
     return ans;
 }
 double dist(point a,point b){
     return sqrt((a-b)*(a-b));
 }
 bool cmp_angle(point a,point b){
     double res=(a-tmp)^(b-tmp);
     if (res) return res>;
     return dist(a,tmp)<dist(b,tmp);
 }
 double largest_empty_convex(point p[],int n){
     double ans=;
     for (int i=;i<n;i++){
         tmp=p[i];
         int cnt=;
         for (int j=;j<n;j++){
             if (p[j].y>tmp.y || p[j].y==tmp.y && p[j].x>tmp.x) pp[cnt++]=p[j];
         }
         sort(pp,pp+cnt,cmp_angle);  //根据极角排序
         ans=max(ans,empty_convex(pp,cnt,tmp));
     }
     return ans;
 }
 int main(){
     int t,n; cin >> t;
     while (t--){
         cin >> n;
         for (int i=;i<n;i++) cin >> p[i].x >> p[i].y;
         double ans=largest_empty_convex(p,n);
         printf("%.1f\n",ans);
     }
     return ;
 } 

poj1259