UVA 10173 最小矩形覆盖（凸包+旋转卡壳）

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传送门

题意正如标题。

可为什么用旋转卡壳呢？

先把有的点用一个凸包维护，那么这个凸包一定覆盖了所有的点，如果要有一个矩形覆盖所有的点，

那么他一定有一条边和凸包的一条边重合。

#include<bits/stdc++.h>#define eps 1e-8using namespace  std;const int _max = 1e3 + 10;const double PI = acos(-1);int n;int sgn(double x) //三态函数{if(fabs(x)<eps) return 0;else return x<0?-1:1;}struct point{double x,y;} p[_max],res[_max];bool mult(point sp,point ep,point op){return (sp.x - op.x) * (ep.y - op.y)       >= (ep.x - op.x) * (sp.y - op.y);}bool operator < (const point &l, const point &r){return l.y < r.y ||(l.y == r.y && l.x < r.x);}int graham(point pnt[],int n, point res[]) //构造凸包{int i,len ,k = 0,top = 1;sort(pnt,pnt+n);if(n == 0)return 0;res[0] = pnt[0];if(n == 1)return 1;res[1] = pnt[1];if(n == 2)return 2;res[2] = pnt[2];for(int i =2; i < n; ++ i){while(top && mult(pnt[i],res[top],res[top-1]))top--;res[++top] = pnt[i];}len = top;res[++top] = pnt[n - 2];for(i = n - 3; i >= 0; -- i){while(top!=len && mult(pnt[i],res[top],res[top-1]))top--;res[++top] = pnt[i];}return top;//返回凸包中点的个数}double len(point A,point B) //返回向量AB的模平方{return (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);}double dot(point A,point B,point C) //点乘{return (C.x-A.x)*(B.x-A.x)+(C.y-A.y)*(B.y-A.y);}double cross(point A,point B,point C) //叉乘{return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);}double minRetangleCover() //最小矩形面积覆盖(旋转卡壳){if(n < 3) return 0.0;res[n] = res [0];double ans = -1;int r = 1, p = 1,q;for(int i = 0; i < n; ++ i){//卡出离边 res[i]-res[i+1]最远的点while(sgn(cross(res[i],res[i+1],res[r+1])-cross(res[i],res[i+1],res[r]))>=0)r = (r+1)% n;//卡出res[i]-res[i+1]方向上正向n最远的点while(sgn(dot(res[i],res[i+1],res[p+1])-dot(res[i],res[i+1],res[p]))>=0)p = (p+1)% n;if(i == 0) q = p;//卡出res[i]-res[i+1]方向上负向最远的点while(sgn(dot(res[i],res[i+1],res[q+1])-dot(res[i],res[i+1],res[q]))<=0)q = (q+1)% n;double d = len(res[i],res[i+1]);double temp = cross(res[i],res[i+1],res[r])*(dot(res[i],res[i+1],res[p])-dot(res[i],res[i+1],res[q]))/d;if(ans < 0 || ans > temp) ans = temp;}return ans;}int main(){#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);#endif // ONLINE_JUDGEwhile(scanf("%d",&n)==1 && n){for(int i = 0; i < n; ++ i){scanf("%lf%lf",&p[i].x,&p[i].y);}n = graham(p,n,res);//构造凸包printf("%.4f\n",minRetangleCover());}return 0;}

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