POJ 2079(计算几何初步——凸包加旋转卡壳)

唉～～～～～这题实在是坑爹，数据范围1<=N<=50000,很明显O(N^2)的算法过不了，但是居然过了-_-!!!!!!!!!!!!

跪求神牛高效算法。。。

先来分析下这题，给出N个点，从中选出3个点，使其面积达到最大。

先用凸包进行初步优化：因为最大面积三角形的顶点一定在凸包上。

然后直接在凸包上暴力枚举是肯定不行的。。O（N^3）

所以这要用到一种思想，旋转卡壳。

能够将其中一个点的枚举次数平摊为O（1）.

所以总的时间复杂度为O(N^2).

旋转卡壳枚举思路：

先固定i,j这2个邻接的顶点。然后找出triangle_area(i,j,k)最大的那个K点。

然后再固定i,枚举j点，由于k点是随着j点的变化在变化，所以K点不必从开头重新枚举。

View Code

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<queue>
  5 #include<vector>
  6 #include<map>
  7 #include<set>
  8 #include<cmath>
  9 #include<complex>
 10 #include<algorithm>
 11 #define EPS 1e-10
 12 #define largty(x,y) (x-EPS)>(y)
 13 #define smalty(x,y) (x+EPS)<(y)
 14 #define nlargty(x,y) (x-EPS)<(y)
 15 #define nsmalty(x,y) (x+EPS)>(y)
 16 #define equal(x,y) (nlargty(x,y) && (nsmalt(x,y)))
 17 //#define x real()
 18 //#define y imag()
 19 #define max(a,b) (a)>(b)?(a):(b)
 20 #define min(a,b) (a)<(b)?(a):(b)
 21 #define MAXN 50004
 22 using namespace std;
 23 //typedef complex<double> point;
 24 struct point {
 25     double x, y;
 26     point(){}
 27     point(double _x, double _y):x(_x),y(_y){}
 28     point(const point &temp)
 29     {
 30         x = temp.x;
 31         y = temp.y;
 32     }
 33 }data[MAXN],convex[MAXN];
 34 point start(0,0);
 35 
 36 inline double cross(const point &p1, const point &p2, const point &q1, const point &q2)
 37 {
 38     return (q2.y - q1.y)*(p2.x - p1.x) - (q2.x - q1.x)*(p2.y - p1.y);    
 39 }
 40 
 41 inline bool cmp1(const point &a, const point &b)
 42 {
 43     return a.y < b.y || (a.y == b.y && a.x < b.x);    
 44 }
 45 
 46 inline bool cmp2(const point &a, const point &b)//逆时针排序 
 47 {
 48     point origin;
 49     origin = start;
 50     return cross(origin,a,origin,b) > 0 || (cross(origin,a,origin,b) == 0 && fabs(a.x) < fabs(b.x));
 51 }
 52 
 53 inline void convex_hull(int &cnt, int n)
 54 {
 55     sort(data,data+n,cmp1);
 56     start = data[0];
 57     sort(data+1,data+n,cmp2);
 58     convex[cnt++] = start;
 59     convex[cnt++] = data[1];
 60     for (int i(2); i<n; ++i) {
 61         while (cnt >= 2 && cross(convex[cnt-2],convex[cnt-1],convex[cnt-1],data[i]) <= 0)--cnt;
 62         convex[cnt++] = data[i];    
 63     }        
 64 }
 65 
 66 inline double rotating_calipers(int n)
 67 {
 68     int j = 1,k = 0;
 69     double area(0);
 70     for (int i(0); i<n; ++i) {
 71         j = (i + 1)%n;
 72         k = (j + 1)%n;
 73         while (fabs(cross(convex[i],convex[j],convex[i],convex[k])) < 
 74                     fabs(cross(convex[i],convex[j],convex[i],convex[(k+1)%n]))) {
 75             k = (k + 1)%n;            
 76         }
 77         while (j != i && k != i) {
 78             area = max(area,fabs(cross(convex[i],convex[j],convex[i],convex[k])));
 79             while (fabs(cross(convex[i],convex[j],convex[i],convex[k])) <
 80                          fabs(cross(convex[i],convex[j],convex[i],convex[(k+1)%n]))) {
 81                 k = (k + 1)%n;                    
 82             }
 83             j = (j + 1)%n;    
 84         }
 85     }
 86     return area;    
 87 }
 88 
 89 int main()
 90 {
 91     int n;
 92     while (scanf("%d",&n) && n != -1) {
 93         for (int i(0); i<n; ++i) {
 94             scanf("%lf%lf",&data[i].x,&data[i].y);    
 95         }
 96         if (n <= 2) {
 97             cout<<"0.00"<<endl;
 98             continue;    
 99         }    
100         int cnt(0);
101         convex_hull(cnt,n);
102         if (cnt <= 2) {
103             cout<<"0.00"<<endl;
104             continue;    
105         } else {
106             if (cnt == 3) {
107                 printf("%.2lf\n",fabs(cross(convex[0],convex[1],convex[0],convex[2]))/2);    
108             } else {
109                 printf("%.2lf\n",rotating_calipers(cnt)/2);    
110             }    
111         }
112     }
113     return 0;    
114 }