uva11168 Airport(凸包)

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uva11168

题目

就是说有很多的居民点，现要建一条直线跑道，使得每个居民点到跑道的平均距离最小。

思路

求凸包，注意1个点和2个点的情况。

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define pi acos (-1)using namespace std;typedef long long ll;const int maxn=10010;const double INF = 1e30;int tot;struct point{    double x, y;    point (double _x = 0, double _y = 0) : x(_x), y(_y) {}    point operator - (point a) const    {        return point (x-a.x, y-a.y);    }    point operator + (point a) const    {        return point (x+a.x, y+a.y);    }    bool operator < (const point &a) const    {        return x < a.x || (x == a.x && y < a.y);    }} p[maxn];const double eps = 1e-10;int dcmp (double x){    if (fabs (x) < eps)        return 0;    else return x < 0 ? -1 : 1;}double cross (point a, point b){    return a.x*b.y-a.y*b.x;}double dot (point a, point b){    return a.x*b.x + a.y*b.y;}double AngleToRad (double x){    return x*pi/180;}point rotate (point a, double rad){    return point (a.x*cos (rad)-a.y*sin (rad), a.x*sin (rad)+a.y*cos (rad));}double ConvexPolygonArea (point *p, int n){    double area = 0;    for (int i = 1; i < n-1; i++)        area += cross (p[i]-p[0], p[i+1]-p[0]);    return area/2.0;}int n, m;double l;point ch[maxn];int ConvexHull (int n){    sort (p, p+n);    int m = 0;    for (int i = 0; i < n; i++)    {        while (m > 1 && cross (ch[m-1]-ch[m-2], p[i]-ch[m-1]) <= 0)            m--;        ch[m++] = p[i];    }    int k = m;    for (int i = n-2; i >= 0; i--)    {        while (m > k && cross (ch[m-1]-ch[m-2], p[i]-ch[m-1]) <= 0)            m--;        ch[m++] = p[i];    }    if (n > 1)        m--;    return m;}double dis (point a, point b){    double xx = a.x-b.x, yy = a.y-b.y;    return sqrt (xx*xx + yy*yy);}int main(){    int T;    scanf("%d",&T);    int kase=1;    while(T--)    {        tot=0;        scanf("%d",&n);        for(int i=0; i<n; i++)        {            double x,y;            scanf("%lf%lf",&x,&y);            p[tot++]=point(x,y);        }        if(n==2||n==1)        {            printf("Case #%d: 0.000\n",kase++);            continue;        }        m=ConvexHull(tot);        double ans=INF;        ch[m+1]=ch[1];        for(int i=1; i<=m; i++)        {            double temp=0;            double dist=dis(ch[i],ch[i+1]);            for(int j=0; j<tot; j++)            {                temp+=cross (ch[i]-p[j], ch[i+1]-p[j])/dist;            }            ans=min(ans,temp);        }        printf("Case #%d: %.3lf\n",kase++,ans/(n*1.0));    }    return 0;}

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