UVa 11168 Airport凸包与直线方程

题意：给出平面上的n个点，求一条直线，使得所有点在该直线的同一侧且所有点到该直线的距离和最小，输出该距离和。

思路:要使所有点在该直线的同一侧，明显是直接利用凸包的边更优。所以枚举凸包的没条边，然后求距离和。直线一般式为Ax + By + C = 0.点(x0, y0)到直线的距离为

fabs(Ax0+By0+C)/sqrt(A*A+B*B).由于所有点在直线的同一侧，那么对于所有点，他们的(Ax0+By0+C)符号相同，显然可以累加出sumX和sumY，然后统一求和。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF 800000000
int sumX, sumY, N;
struct Point{int x, y;};
Point p[10002], ch[10002];
bool cmp(Point a, Point b){
    return a.x == b.x ? a.y < b.y : a.x < b.x;
}
int Cross(Point a, Point b, Point c){
    return (a.x-c.x)*(b.y-c.y) - (a.y-c.y)*(b.x-c.x);
}
const double eps = 1e-10;
int Dcmp(double x){
    if(x < eps)return 0;
    return x < 0 ? -1 : 1;
}
int ConvexHull()//凸包
{
    sort(p, p+N, cmp);
    int m = 0;
    for(int i = 0; i < N; i++){
        while(m > 1 && Dcmp(Cross(ch[m-1], p[i], ch[m-2])) <= 0)m--;
        ch[m++] = p[i];
    }
    int k = m;
    for(int i = N-2; i >= 0; i--){
        while(m > k && Dcmp(Cross(ch[m-1], p[i], ch[m-2])) <= 0)m--;
        ch[m++] = p[i];
    }
    if(N > 1)m--;
    return m;
}
double getDist(Point a, Point b){
    double A, B, C, k, v;
    A = a.y - b.y;
    B = b.x - a.x;
    C = a.x*b.y - a.y*b.x;
    k = fabs(A*sumX + B*sumY + N * C);
    v = sqrt(A * A + B * B);
    return k / v;
}
double Line(int m)
{
    double minL = INF*1.0, L;
    for(int i = 0; i < m; i++){
        L = getDist(ch[i], ch[(i+1)%m]);//得到所有点到凸包一条边的距离
        minL = min(minL, L);
    }
    return minL;
}
int main()
{
    int i, j, t;
    cin>>t;
    int cas = 1;
    while(t--)
    {
        cin>>N;
        sumX = sumY = 0;
        for(i = 0; i < N; i++){
            scanf("%d%d", &p[i].x, &p[i].y);
            sumX += p[i].x;
            sumY += p[i].y;
        }
        printf("Case #%d: ", cas++);
        if(N == 1 || N== 2){
            printf("0.000
");
            continue;
        }
        printf("%.3lf
", Line( ConvexHull() ) / (N*1.0));
    }
    return 0;
}