1393: Robert Hood 旋转卡壳 凸包

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1393

http://poj.org/problem?id=2187 Beauty Contest

1393: Robert Hood

 

Description

Input

Output

Sample Input

5
-4 1
-100 0
0 4
2 -3
2 300

Sample Output

316.86590223

HINT

 

 

Source

 

 

分析：

给你 N 个点, 求所有点中最远两点距离。即是凸包直径。

凸包+旋转卡壳 

 

AC代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn = 100000+10;
int n,m;

struct Point{
    double x,y;
    Point(){};
    Point(double _x, double _y)
    {
        x = _x;
        y = _y;
    }

    Point operator - (const Point & B) const
    {
        return Point(x-B.x, y-B.y);
    }
}p[maxn], ch[maxn];

bool cmp(Point p1, Point p2)
{
    if(p1.x == p2.x) return p1.y < p2.y;
    return p1.x < p2.x;
}

int squarDist(Point A, Point B) /**距离的平方*/
{
    return (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
}

double Cross(Point A, Point B) /**叉积*/
{
    return A.x*B.y-A.y*B.x;
}

void ConvexHull() /** 基于水平的Andrew算法求凸包 */
{
    sort(p,p+n,cmp); /**先按照 x 从小到大排序, 再按照 y 从小到大排序*/
    m = 0;

    for(int i = 0; i < n; i++) /** 从前往后找 */
    {
        while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    int k = m;
    for(int i = n-2; i >= 0; i--) /**从后往前找, 形成完整的封闭背包*/
    {
        while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    if(n > 1) m--;
}

int rotating_calipers() /**旋转卡壳模板*/
{
    int q = 1;
    int ans = 0;
    ch[m] = ch[0]; /**凸包边界处理*/
    for(int i = 0; i < m; i++) /**依次用叉积找出凸包每一条边对应的最高点*/
    {
        while(Cross(ch[i+1]-ch[i], ch[q+1]-ch[i]) > Cross(ch[i+1]-ch[i], ch[q]-ch[i]))
            q = (q+1)%m;
        ans = max(ans, max(squarDist(ch[i], ch[q]), squarDist(ch[i+1], ch[q+1])));
    }
    return ans;
}

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) break;
        for(int i = 0; i < n; i++)
            scanf("%lf%lf", &p[i].x, &p[i].y);

        ConvexHull();

        printf("%.8lf\n", sqrt(rotating_calipers()));
    }
    return 0;
}

 

 

同学用结构体 + 遍历凸包也可以解决。

 

AC代码：

#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
struct point
{
    int x;
    int y;
}p[100005],res[100005];
int cmp(point p1,point p2)
{
    return p1.y<p2.y||(p1.x==p2.x&&p1.x<p2.x);
}
bool ral(point p1,point p2,point p3)
{
    return (p2.x-p1.x)*(p3.y-p1.y)>(p3.x-p1.x)*(p2.y-p1.y);
}
int main()
{
    int n,i,j;
    while((scanf("%d",&n))!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%d %d",&p[i].x,&p[i].y);
        sort(p,p+n,cmp);
        res[0]=p[0];
        res[1]=p[1];
        int top=1;
        for(i=2;i<n;i++)
        {
            while(top&&!ral(res[top],res[top-1],p[i]))
                top--;
            res[++top]=p[i];
        }
        int len=top;
        res[++top]=p[n-2];
        for(i=n-3;i>=0;i--)
        {
            while(top!=len&&!ral(res[top],res[top-1],p[i]))
                top--;
            res[++top]=p[i];
        }
        double maxx=0;
        for(i=0;i<top;i++)
        {
            for(j=i+1;j<top;j++)
            {
                double s=(res[i].x-res[j].x)*(res[i].x-res[j].x)+(res[i].y-res[j].y)*(res[i].y-res[j].y);
                if(s>maxx)
                    maxx=s;
            
            }
        }
        printf("%.8lf\n",sqrt(maxx));
    }
    return 0;
}