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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<complex>
#include<algorithm>
#define EPS 1e-10
#define largty(x,y) (x-EPS)>(y)
#define smalty(x,y) (x+EPS)<(y)
#define nlargty(x,y) (x-EPS)<(y)
#define nsmalty(x,y) (x+EPS)>(y)
#define equal(x,y) (nlargty(x,y) && (nsmalt(x,y)))
//#define x real()
//#define y imag()
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
#define MAXN 50004
using namespace std;
//typedef complex<double> point;
struct point {
double x, y;
point(){}
point(double _x, double _y):x(_x),y(_y){}
point(const point &temp)
{
x = temp.x;
y = temp.y;
}
}data[MAXN],convex[MAXN];
point start(0,0);

inline double cross(const point &p1, const point &p2, const point &q1, const point &q2)
{
return (q2.y - q1.y)*(p2.x - p1.x) - (q2.x - q1.x)*(p2.y - p1.y);
}

inline bool cmp1(const point &a, const point &b)
{
return a.y < b.y || (a.y == b.y && a.x < b.x);
}

inline bool cmp2(const point &a, const point &b)//逆时针排序
{
point origin;
origin = start;
return cross(origin,a,origin,b) > 0 || (cross(origin,a,origin,b) == 0 && fabs(a.x) < fabs(b.x));
}

inline void convex_hull(int &cnt, int n)
{
sort(data,data+n,cmp1);
start = data[0];
sort(data+1,data+n,cmp2);
convex[cnt++] = start;
convex[cnt++] = data[1];
for (int i(2); i<n; ++i) {
while (cnt >= 2 && cross(convex[cnt-2],convex[cnt-1],convex[cnt-1],data[i]) <= 0)--cnt;
convex[cnt++] = data[i];
}
}

inline double rotating_calipers(int n)
{
int j = 1,k = 0;
double area(0);
for (int i(0); i<n; ++i) {
j = (i + 1)%n;
k = (j + 1)%n;
while (fabs(cross(convex[i],convex[j],convex[i],convex[k])) <
fabs(cross(convex[i],convex[j],convex[i],convex[(k+1)%n]))) {
k = (k + 1)%n;
}
while (j != i && k != i) {
area = max(area,fabs(cross(convex[i],convex[j],convex[i],convex[k])));
while (fabs(cross(convex[i],convex[j],convex[i],convex[k])) <
fabs(cross(convex[i],convex[j],convex[i],convex[(k+1)%n]))) {
k = (k + 1)%n;
}
j = (j + 1)%n;
}
}
return area;
}

int main()
{
int n;
while (scanf("%d",&n) && n != -1) {
for (int i(0); i<n; ++i) {
scanf("%lf%lf",&data[i].x,&data[i].y);
}
if (n <= 2) {
cout<<"0.00"<<endl;
continue;
}
int cnt(0);
convex_hull(cnt,n);
if (cnt <= 2) {
cout<<"0.00"<<endl;
continue;
} else {
if (cnt == 3) {
printf("%.2lf\n",fabs(cross(convex[0],convex[1],convex[0],convex[2]))/2);
} else {
printf("%.2lf\n",rotating_calipers(cnt)/2);
}
}
}
return 0;
}