旋转卡壳解决一类问题（程序）——未完待续

1、旋转卡壳求凸多边形直径（即求平面点集最远点对）

http://poj.org/problem?id=2187   本题数据较水，可以直接暴力

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<cstdlib>#include<queue>#include<stack>#include<map>#include<vector>#include<algorithm>#include<ctime>using namespace std;#define maxn 50005 #define eps 1e-6#define pi acos(-1.0)int Fabs(double d){if(fabs(d)<eps) return 0;else return d>0?1:-1;}struct point {double x,y;bool operator == (const point &p){return Fabs(x-p.x)==0&&Fabs(y-p.y)==0;}}p[maxn],sta[maxn];int top;double x_multi(point p1,point p2,point p3)    {      return (p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y);  } double Dis(point p1,point p2){return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}bool cmp(point a,point b){if(Fabs(x_multi(p[0],a,b))>0) return 1;if(Fabs(x_multi(p[0],a,b))<0) return 0;if(Fabs(Dis(p[0],a)-Dis(p[0],b))<0)return 1;return 0;}void Graham(int n){int i,k=0,tot;for(i=1;i<n;i++)if((p[i].y<p[k].y)||((p[i].y==p[k].y)&&(p[i].x<p[k].x)))k=i;swap(p[0],p[k]);sort(p+1,p+n,cmp);tot=1;    for(i=2;i<n;i++)        if (Fabs(x_multi(p[i],p[i-1],p[0])))            p[tot++]=p[i-1];    p[tot++]=p[n-1];sta[0]=p[0],sta[1]=p[1];i=top=1;for(i=2;i<tot;i++){         while(top>=1&&Fabs(x_multi(p[i],sta[top],sta[top-1]))>=0){             if(top==0) break;             top--;         }         sta[++top]=p[i];     }   }int main(){int n,i,j;while(~scanf("%d",&n)){for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);double ans=0.0,tmp;Graham(n);for(i=0;i<=top;i++)for(j=i+1;j<=top;j++){tmp=Dis(sta[i],sta[j]);if(tmp>ans) ans=tmp;}/*旋转卡壳*//*sta[top+1]=sta[0];for(i=0,j=1;i<=top;i++){while(x_multi(sta[i],sta[i+1],sta[j+1])>x_multi(sta[i],sta[i+1],sta[j]))j=(j+1)%(top+1);ans=max(ans,max(Dis(sta[i+1],sta[j+1]),Dis(sta[i],sta[j])));}*/printf("%.0lf\n",ans*ans);}return 0;}

2、旋转卡壳求凸多边形的宽

问题引入，在一个二维平面上， 一个凸多边形的物体，要经过一个隧道， 隧道建造的价值取决于隧道的宽度， 现在求隧道的最小价值。

2011年world final一到标配题：http://livearchive.onlinejudge.org/external/51/5138.pdf

提交地址：http://acm.hust.edu.cn:8080/judge/problem/viewProblem.action?id=19848

数据http://cm.baylor.edu/ICPCWiki/Wiki.jsp?page=Problem%20Resources

思路参考：http://www.ipplelife.com/?p=615

code: 省略graham()部分：

double ptoline(point a,point b,point d){if(a.x==b.x) //无斜率return fabs(d.x-a.x);double k=(a.y-b.y)/(a.x-b.x);double tb=a.y-k*a.x;return fabs(k*d.x-d.y+tb)/sqrt(k*k+1);}int main(){int n,i,cas=0;while(scanf("%d",&n),n){for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);Graham(n);double ans=1e+10;//最好初始无穷大sta[top+1]=sta[0];int q=1;for(i=0;i<=top;i++){while(fabs(x_multi(sta[q+1],sta[i],sta[i+1]))>fabs(x_multi(sta[q],sta[i],sta[i+1])))q=(q+1)%(top+1);ans=min(ans,ptoline(sta[i],sta[i+1],sta[q]));}printf("Case %d: %.2lf\n",++cas,ceil(ans*100)/100);}return 0;}