ural 1588 Jamaica

思路：枚舉所有點，計算兩點長度，同時在搜索所有點，判斷是否有共線的，有的話，只算距離最長的就可以了。對於多個點在同一直線上，最長距離肯定是兩端的點，所以需要預排序。時間複雜度O(n*n*n).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

#define eps 1e-8
const int MAX = 309;
const int INF = 1<<30;
struct Point {
    double x, y;
};

Point p[MAX];
bool flag[MAX][MAX];

double direction(int i, int j, int k)
{
    return ((p[j].x-p[i].x)*(p[k].y-p[i].y)-(p[j].y-p[i].y)*(p[k].x-p[i].x));
}

double getdis(int i, int j)
{
    double dx,dy;
    dx = p[j].x - p[i].x;
    dy = p[j].y - p[i].y;
    return (sqrt(dx*dx + dy*dy));
}
double direction(Point a, Point b)
{
    return ((a.x-p[0].x)*(b.y-p[0].y) - (a.y-p[0].y)*(b.x-p[0].x));
}

double getdis(Point a)
{
    double dx,dy;
    dx = a.x - p[0].x;
    dy = a.y - p[0].y;
    return (sqrt(dx*dx + dy*dy));
}

int cmp(Point a, Point b)
{
    if(fabs(direction(a, b)) < eps)
        return getdis(a) < getdis(b);
    else if(direction(a, b) < 0)
        return 1;
    else
        return 0;
}

int main()
{
    int i, j, k, n;
    double ans = 0, ma;
    scanf("%d", &n);
    k = 0;
    for (i = 0; i < n; ++i) {
        scanf("%lf %lf", &p[i].x, &p[i].y);
        if ((p[i].x == p[k].x && p[i].y < p[k].y) || (p[i].x < p[k].x))
            k = i;
    }
    swap(p[0], p[k]);
    sort(p+1, p+n, cmp);
    memset(flag, 0, sizeof(flag));
    for (i = 0; i < n; ++i)
        for (j = i+1; j < n; ++j) {
                ma = getdis(i, j);
                for (k = j + 1; k < n; ++k) {
                    if(fabs(direction(i, j, k)) < eps){
                        ma = max(ma, getdis(i, k));
                        ma = max(ma, getdis(j, k));
                        flag[i][k] = flag[k][i] = 1;
                        flag[j][k] = flag[k][j] = 1;
                    }
                }
                if (!flag[i][j]) {
                    ans += ma;
                    flag[i][j] = 1;
                    flag[j][i] = 1;
                }
            }

    printf("%.0lf\n", ans);
    return 0;
}