Enum:Game of Lines(POJ 3668)

　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　　　画直线

　　题目大意：给定一些点集，要你找两点之间的连线不平行的有多少条

　　数据量比较少，直接暴力枚举，然后放到set查找即可

#include <iostream>
#include <functional>
#include <algorithm>
#include <set>

using namespace std;

static struct _p_set
{
    long double x, y;
}points[201];
set<long double>lines;

int gcd(const int, const int);

int main(void)
{
    int point_sum, cut;
    long double tmp;
    //pair<int,int>tmp;
    while (~scanf("%d", &point_sum))
    {
        lines.clear();
        for (int i = 0; i < point_sum; i++)
            scanf("%lf%lf", &points[i].x, &points[i].y);
        for (int i = 0; i < point_sum; i++)
        {
            for (int j = i + 1; j < point_sum; j++)
            {
                if ((points[i].x - points[j].x) != 0)
                    tmp = (points[i].y - points[j].y) / (points[i].x - points[j].x);
                else
                    tmp = (long double)INT_MAX;
                lines.insert(tmp);
            }
        }
        printf("%d
", lines.size());
    }

    return EXIT_SUCCESS;
}

　　

　　在讨论版那里还找到了一种很新奇的做法，可以无视除数是0和精度的问题

#include <iostream>
#include <functional>
#include <algorithm>
#include <set>

using namespace std;

static struct _p_set
{
    int x, y;
}points[201];
set<pair<int,int>>lines;

int gcd(const int, const int);

int main(void)
{
    int point_sum, cut;
    pair<int,int>tmp;
    while (~scanf("%d", &point_sum))
    {
        lines.clear();
        for (int i = 0; i < point_sum; i++)
            scanf("%d%d", &points[i].x, &points[i].y);
        for (int i = 0; i < point_sum; i++)
        {
            for (int j = i + 1; j < point_sum; j++)
            {
                cut = gcd(points[i].y - points[j].y, points[i].x - points[j].x);
                tmp.first = (points[i].y - points[j].y) / cut;
                tmp.second = (points[i].x - points[j].x) / cut;
                lines.insert(tmp);
            }
        }
        printf("%d
", lines.size());
    }
    return EXIT_SUCCESS;
}

int gcd(const int a, const int b)
{
    if (b == 0)
        return a;
    return gcd(b, a%b);
}

　　

　　其实时间差不多