Cycling Roads

几何

When Vova was in Shenzhen, he rented a bike and spent most of the time cycling around the city. Vova was approaching one of the city parks when he noticed the park plan hanging opposite the central entrance. The plan had several marble statues marked on it. One of such statues stood right there, by the park entrance. Vova wanted to ride in the park on the bike and take photos of all statues. The park territory has multiple bidirectional cycling roads. Each cycling road starts and ends at a marble statue and can be represented as a segment on the plane. If two cycling roads share a common point, then Vova can turn on this point from one road to the other. If the statue stands right on the road, it doesn’t interfere with the traffic in any way and can be photoed from the road.
Can Vova get to all statues in the park riding his bike along cycling roads only?
Input
The first line contains integers n and m that are the number of statues and cycling roads in the park (1 ≤ m < n ≤ 200) . Then n lines follow, each of them contains the coordinates of one statue on the park plan. The coordinates are integers, their absolute values don’t exceed 30 000. Any two statues have distinct coordinates. Each of the following m lines contains two distinct integers from 1 to n that are the numbers of the statues that have a cycling road between them.
Output
Print “YES” if Vova can get from the park entrance to all the park statues, moving along cycling roads only, and “NO” otherwise.
Example
input output

题目大意：给出n个点，m条线段，问所有的点是否可以相互到达。
给出的点有两个变量，x,y。给出的线段，有两个变量，每一条线段代表第一个点和第二个点之间有一条线段。

可以使用并查集+线段，首先，给出线段之后，把在线段上的点都合并，然后遍历看看这条线段是否与其他线段相交，如果相交，则合并其他点。如果最后只有一个祖先，则是YES，否则，NO。

#include <bits/stdc++.h>
using namespace std;
int pre[322];
int ans;
const double pi = acos(-1.0);
const double eps = 1e-8;
int Find(int a)
{
    if(a!=pre[a])
        return pre[a] = Find(pre[a]);
    return a;
}
void Merge(int a, int b)
{
    int aa = Find(a);
    int bb = Find(b);
    if(aa!=bb)
    {
        pre[aa] = bb;
        ans--;
    }
}
int cmp(double x)
{
    if(fabs(x)<eps)
        return 0;
    if(x>0)
        return 1;
    return -1;
}
struct point
{
    double x, y;
    int id;
    point(){}
    point(double a, double b)
    {
        x = a;
        y = b;
    }
    friend point operator + (const point &a, const point &b)
    {
        return point (a.x + b.x, a.y + b.y);
    }
    friend point operator - (const point &a, const point &b)
    {
        return point (a.x - b.x, a.y - b.y);
    }
}Point[345];
double dot(const point &a, const point &b)//点积
{
    return a.x*b.x+a.y*b.y;
}
double det(const point &a, const point &b)//叉积
{
    return a.x*b.y - a.y*b.x;
}
struct line
{
    point a, b;
    line(){};
    line(point x, point y):a(x), b(y){}
}Line[322];
line point_make_line(const point a, const point b)
{
    return line(a, b);
}
//bool PointOnSegment (point p, point s, point t)
//{
//    return cmp(det(p-s, t-s))==0&&cmp(dot(p-s, p-t))<=0;
//}
bool PointOnSegment (point p, line A)
{
    return cmp(det(A.a - p, A.b-p))==0&&cmp(dot(A.a-p, A.b-p))<=0;
}
//bool parallel(line a, line b)
//{
//    return !cmp(det(a.a-a.b, b.a-b.b));
//}
//bool line_make_point(line a, line b)
//{
//    if(parallel(a, b))
//        return false;
//    return true;
//}
bool line_make_point(line l1, line l2)
{
    if (PointOnSegment(l1.a, l2) || PointOnSegment(l1.b, l2) || PointOnSegment(l2.a, l1) || PointOnSegment(l2.b, l1))
    {
        return true;
    }
    double c1 = det(l1.a - l2.a, l1.a - l1.b);
    double c2 = det(l1.a - l2.b, l1.a - l1.b);
    double c3 = det(l2.a - l1.a, l2.a - l2.b);
    double c4 = det(l2.a - l1.b, l2.a - l2.b);
    return cmp(c1) * cmp(c2) < 0 && cmp(c3) * cmp(c4) < 0;
}
int main()
{
    int n, m;
    cin>>n>>m;
    ans = n;
    for(int i=1;i<=n;i++)
    {
        cin>>Point[i].x>>Point[i].y;
        Point[i].id = i;
    }
    for(int i=1;i<=n;i++)
    {
        pre[i] = i;
    }
    for(int i=1;i<=m;i++)
    {
        int a, b;
        cin>>a>>b;
        Line[i] = point_make_line(Point[a], Point[b]);
        for(int j=1;j<=n;j++)
        {
//            if(PointOnSegment(Point[j], Point[a], Point[b]))
//                {
//                    Merge(Point[j].id, Point[a].id);
//                    Merge(Point[j].id, Point[b].id);
//                }
                if(PointOnSegment(Point[j], Line[i]))
                {
                    Merge(Point[j].id, Line[i].a.id);
                    Merge(Point[j].id, Line[i].b.id);
                }
        }
        for(int j=1;j<i;j++)
        {
            if(line_make_point(Line[i], Line[j]))
            {
                Merge(Line[i].a.id, Line[j].a.id);
                Merge(Line[i].a.id, Line[j].b.id);
                Merge(Line[i].b.id, Line[j].a.id);
                Merge(Line[i].b.id, Line[j].b.id);
            }
        }
    }
    if(ans==1)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
    return 0;
}

几何

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