旋转卡壳
1. Luogu P1452 [USACO03FALL]Beauty Contest G /【模板】旋转卡壳
思路：距离最远的点一定是凸壳上的两点
双指针枚举，i指针枚举凸壳的边，j指针在前面枚举最远点，优选答案
注意，两个指针都是向前走的，保证旋转卡壳时间为O(n)
时间：O(n*logn + n)
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 6 #define N 50010 7 #define x first 8 #define y second 9 #define Point pair<int,int> 10 Point p[N],s[N]; 11 int n; 12 13 int cross(Point a,Point b,Point c){ //叉积 14 return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x); 15 } 16 int dis(Point a, Point b){ //距离平方 17 return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); 18 } 19 void Andrew(){ 20 sort(p+1,p+n+1); 21 int t=0; 22 for(int i=1; i<=n; i++){ //下凸包 23 while(t>1&&cross(s[t-1],s[t],p[i])<=0)t--; 24 s[++t]=p[i]; 25 } 26 int k=t; 27 for(int i=n-1; i>=1; i--){ //上凸包 28 while(t>k&&cross(s[t-1],s[t],p[i])<=0)t--; 29 s[++t]=p[i]; 30 } 31 n=t-1; //n为凸包上的点数 32 } 33 int rotating_calipers(){ //旋转卡壳 34 int res=0; 35 for(int i=1,j=2; i<=n; i++){ 36 while(cross(s[i],s[i+1],s[j])<cross(s[i],s[i+1],s[j+1]))j=j%n+1; 37 res=max(res,max(dis(s[i],s[j]),dis(s[i+1],s[j]))); 38 } 39 return res; 40 } 41 int main(){ 42 scanf("%d",&n); 43 for(int i=1; i<=n; i++) scanf("%d%d",&p[i].x,&p[i].y); 44 Andrew(); 45 printf("%d\n",rotating_calipers()); 46 return 0; 47 }
2. Luogu P4166 [SCOI2007]最大土地面积
思路：内接四边形的对角线一定是凸包的对角线
枚举凸包的对角线，旋转卡壳求最远点a,b，更新答案
时间：O(n*logn + n*n)
1 #include<cstring> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 7 const double eps=1e-8; 8 const double PI=acos(-1); 9 #define N 50010 10 #define x first 11 #define y second 12 #define Point pair<int,int> 13 Point p[N],s[N]; 14 int n; 15 16 Point operator+(Point a, Point b){ //向量+ 17 return {a.x+b.x, a.y+b.y}; 18 } 19 Point operator-(Point a, Point b){ //向量- 20 return {a.x-b.x, a.y-b.y}; 21 } 22 Point operator*(Point a, double p){ //数乘 23 return {a.x*p, a.y*p}; 24 } 25 double cross(Point a,Point b,Point c){ //叉积 26 return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x); 27 } 28 double dot(Point a,Point b,Point c){ //点积 29 return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y); 30 } 31 double dis(Point a,Point b){ //距离 32 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 33 } 34 Point rotate(Point a, double b){ //逆转角 35 return {a.x*cos(b)-a.y*sin(b), a.x*sin(b)+a.y*cos(b)}; 36 } 37 bool cmp(Point a, Point b){ //找最低点 38 return fabs(a.y-b.y)>eps ? a.y<b.y : a.x<b.x; 39 } 40 void zero(Point &a){ //避免-0.00000 41 if(fabs(a.x)<=eps) a.x=0; 42 if(fabs(a.y)<=eps) a.y=0; 43 } 44 void Andrew(){ 45 sort(p+1,p+n+1); 46 int t=0; 47 for(int i=1; i<=n; i++){ //下凸包 48 while(t>1 && cross(s[t-1],s[t],p[i])<=0)t--; 49 s[++t]=p[i]; 50 } 51 for(int k=t, i=n-1; i>=1; i--){ //上凸包 52 while(t>k && cross(s[t-1],s[t],p[i])<=0)t--; 53 s[++t]=p[i]; 54 } 55 n=t-1; //n为凸包上的点数 56 } 57 void rotating_calipers(){ //旋转卡壳 58 double ans=1e20; 59 int a,b,c; a=b=2; //上a,右b,左c 60 for(int i=1;i<=n;i++){ 61 while(cross(s[i],s[i+1],s[a])<cross(s[i],s[i+1],s[a+1]))a=a%n+1; 62 while(dot(s[i],s[i+1],s[b])<dot(s[i],s[i+1],s[b+1]))b=b%n+1; 63 if(i==1)c=a; 64 while(dot(s[i+1],s[i],s[c])<dot(s[i+1],s[i],s[c+1]))c=c%n+1; 65 double d=dis(s[i],s[i+1]); 66 double H=cross(s[i],s[i+1],s[a])/d; 67 double R=dot(s[i],s[i+1],s[b])/d; 68 double L=dot(s[i+1],s[i],s[c])/d; 69 if(ans>(R+L-d)*H){ 70 ans=(R+L-d)*H; 71 p[1]=s[i+1]+(s[i]-s[i+1])*(L/d); 72 p[2]=s[i]+(s[i+1]-s[i])*(R/d); 73 p[3]=p[2]+rotate(s[i+1]-s[i],PI/2)*(H/d); 74 p[4]=p[1]+rotate(s[i+1]-s[i],PI/2)*(H/d); 75 } 76 } 77 printf("%.5lf\n",ans); 78 int k=1; 79 for(int i=2;i<=4;i++) if(cmp(p[i],p[k]))k=i; //找最低点 80 for(int i=1; i<=4; i++){ 81 zero(p[k]); 82 printf("%.5lf %.5lf\n",p[k].x,p[k].y); 83 k=k%4+1; 84 } 85 } 86 int main(){ 87 scanf("%d",&n); 88 for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); 89 Andrew(); 90 rotating_calipers(); 91 return 0; 92 }
3. Luogu P3187 [HNOI2007]最小矩形覆盖
思路：最小矩形的某条边一定在凸壳上的某条边上
枚举凸壳的边，旋转卡壳求三点
时间：O(n*logn + n)
1 #include<cstring> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 7 const double eps=1e-8; 8 const double PI=acos(-1); 9 #define N 50010 10 #define x first 11 #define y second 12 #define Point pair<int,int> 13 Point p[N],s[N]; 14 int n; 15 16 Point operator+(Point a, Point b){ //向量+ 17 return {a.x+b.x, a.y+b.y}; 18 } 19 Point operator-(Point a, Point b){ //向量- 20 return {a.x-b.x, a.y-b.y}; 21 } 22 Point operator*(Point a, double p){ //数乘 23 return {a.x*p, a.y*p}; 24 } 25 double cross(Point a,Point b,Point c){ //叉积 26 return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x); 27 } 28 double dot(Point a,Point b,Point c){ //点积 29 return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y); 30 } 31 double dis(Point a,Point b){ //距离 32 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 33 } 34 Point rotate(Point a, double b){ //逆转角 35 return {a.x*cos(b)-a.y*sin(b), a.x*sin(b)+a.y*cos(b)}; 36 } 37 bool cmp(Point a, Point b){ //找最低点 38 return fabs(a.y-b.y)>eps ? a.y<b.y : a.x<b.x; 39 } 40 void zero(Point &a){ //避免-0.00000 41 if(fabs(a.x)<=eps) a.x=0; 42 if(fabs(a.y)<=eps) a.y=0; 43 } 44 void Andrew(){ 45 sort(p+1,p+n+1); 46 int t=0; 47 for(int i=1; i<=n; i++){ //下凸包 48 while(t>1 && cross(s[t-1],s[t],p[i])<=0)t--; 49 s[++t]=p[i]; 50 } 51 for(int k=t, i=n-1; i>=1; i--){ //上凸包 52 while(t>k && cross(s[t-1],s[t],p[i])<=0)t--; 53 s[++t]=p[i]; 54 } 55 n=t-1; //n为凸包上的点数 56 } 57 void rotating_calipers(){ //旋转卡壳 58 double ans=1e20; 59 int a,b,c; a=b=2; //上a,右b,左c 60 for(int i=1;i<=n;i++){ 61 while(cross(s[i],s[i+1],s[a])<cross(s[i],s[i+1],s[a+1]))a=a%n+1; 62 while(dot(s[i],s[i+1],s[b])<dot(s[i],s[i+1],s[b+1]))b=b%n+1; 63 if(i==1)c=a; 64 while(dot(s[i+1],s[i],s[c])<dot(s[i+1],s[i],s[c+1]))c=c%n+1; 65 double d=dis(s[i],s[i+1]); 66 double H=cross(s[i],s[i+1],s[a])/d; 67 double R=dot(s[i],s[i+1],s[b])/d; 68 double L=dot(s[i+1],s[i],s[c])/d; 69 if(ans>(R+L-d)*H){ 70 ans=(R+L-d)*H; 71 p[1]=s[i+1]+(s[i]-s[i+1])*(L/d); 72 p[2]=s[i]+(s[i+1]-s[i])*(R/d); 73 p[3]=p[2]+rotate(s[i+1]-s[i],PI/2)*(H/d); 74 p[4]=p[1]+rotate(s[i+1]-s[i],PI/2)*(H/d); 75 } 76 } 77 printf("%.5lf\n",ans); 78 int k=1; 79 for(int i=2;i<=4;i++) if(cmp(p[i],p[k]))k=i; //找最低点 80 for(int i=1; i<=4; i++){ 81 zero(p[k]); 82 printf("%.5lf %.5lf\n",p[k].x,p[k].y); 83 k=k%4+1; 84 } 85 } 86 int main(){ 87 scanf("%d",&n); 88 for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); 89 Andrew(); 90 rotating_calipers(); 91 return 0; 92 }
练习题
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 6 #define N 50010 7 #define x first 8 #define y second 9 #define Point pair<int,int> 10 Point p[N],s[N]; 11 int n,top; 12 13 int cross(Point a,Point b,Point c){ //叉积 14 return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x); 15 } 16 void Andrew(){ 17 sort(p+1,p+n+1); 18 top=0; 19 for(int i=1; i<=n; i++){ //下凸包 20 while(top>1&&cross(s[top-1],s[top],p[i])<=0)top--; 21 s[++top]=p[i]; 22 } 23 int t=top; 24 for(int i=n-1; i>=1; i--){ //上凸包 25 while(top>t&&cross(s[top-1],s[top],p[i])<=0)top--; 26 s[++top]=p[i]; 27 } 28 n=top-1; //n为凸包上的点数 29 } 30 int rotating_calipers(){ //旋转卡壳 31 int res=0; 32 for(int i=1; i<=n; i++){ 33 int k=i+1; //k为j到i之间的点 34 for(int j=i+1; j<=n; j++){ 35 while(cross(s[i],s[j],s[k+1])>cross(s[i],s[j],s[k]))k=k%n+1; 36 res=max(res,cross(s[i],s[j],s[k])); 37 } 38 } 39 return res; 40 } 41 int main(){ 42 while(scanf("%d",&n),n!=-1){ 43 for(int i=1; i<=n; i++) scanf("%d%d",&p[i].x,&p[i].y); 44 Andrew(); 45 printf("%.2f\n",rotating_calipers()/2.); 46 } 47 return 0; 48 }
