URAL 1966 Cycling Roads 计算几何

Cycling Roads

题目连接：

http://acm.hust.edu.cn/vjudge/contest/123332#problem/F

Description

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.

Sample Input

4 2

0 0

1 0

1 1

0 1

1 3

4 2

Sample Output

YES

Hint

题意

平面给你n个点，以及m对直线，问你这m条直线是否能够使得所有点都在一个连通块内

题解：

用并查集去维护就好了，如果两条直线相交，就把直线端点的压进并查集就好了。

然后最后统计一下并查集的大小。

代码

#include<bits/stdc++.h>
using namespace std;
/* 常用的常量定义 */
const double INF  = 1E200;
const double EP  = 1E-10;
const int  MAXV = 300;
const double PI  = 3.14159265;
const int maxn = 300;
/* 基本几何结构 */
struct POINT
{
 double x;
 double y;
 POINT(double a=0, double b=0) { x=a; y=b;} //constructor
};
struct LINESEG
{
 POINT s;
 POINT e;
 int a,b;
 LINESEG(POINT a, POINT b) { s=a; e=b;}
 LINESEG() { }
};
struct LINE           // 直线的解析方程 a*x+b*y+c=0  为统一表示，约定 a >= 0
{
   double a;
   double b;
   double c;
   LINE(double d1=1, double d2=-1, double d3=0) {a=d1; b=d2; c=d3;}
};
double multiply(POINT sp,POINT ep,POINT op)
{
 return((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}
// 如果线段u和v相交(包括相交在端点处)时，返回true
//
//判断P1P2跨立Q1Q2的依据是：( P1 - Q1 ) × ( Q2 - Q1 ) * ( Q2 - Q1 ) × ( P2 - Q1 ) >= 0。
//判断Q1Q2跨立P1P2的依据是：( Q1 - P1 ) × ( P2 - P1 ) * ( P2 - P1 ) × ( Q2 - P1 ) >= 0。
bool intersect(LINESEG u,LINESEG v)
{
 return( (max(u.s.x,u.e.x)>=min(v.s.x,v.e.x))&&                     //排斥实验
   (max(v.s.x,v.e.x)>=min(u.s.x,u.e.x))&&
   (max(u.s.y,u.e.y)>=min(v.s.y,v.e.y))&&
   (max(v.s.y,v.e.y)>=min(u.s.y,u.e.y))&&
   (multiply(v.s,u.e,u.s)*multiply(u.e,v.e,u.s)>=0)&&         //跨立实验
   (multiply(u.s,v.e,v.s)*multiply(v.e,u.e,v.s)>=0));
}
/******************************************************************************
判断点p是否在线段l上
条件：(p在线段l所在的直线上) && (点p在以线段l为对角线的矩形内)
*******************************************************************************/
bool online(LINESEG l,POINT p)
{
 return( (multiply(l.e,p,l.s)==0) &&( ( (p.x-l.s.x)*(p.x-l.e.x)<=0 )&&( (p.y-l.s.y)*(p.y-l.e.y)<=0 ) ) );
}
int fa[maxn];
int fi(int u){
    return u != fa[u] ? fa[u] = fi( fa[u] ) : u;
}

void uni(int u ,int v){
    int p1 = fi( u ) , p2 = fi( v );
    if( p1 != p2 ) fa[p1] = p2;
}

POINT p[maxn];
LINESEG L[maxn];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%lf%lf",&p[i].x,&p[i].y);
        fa[i]=i;
    }
    for(int i=1;i<=m;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        L[i].s=p[x],
        L[i].e=p[y];
        L[i].a=x;
        L[i].b=y;
        uni(x,y);
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(online(L[j],p[i])){
                uni(i,L[j].a);
                uni(i,L[j].b);
            }
        }
    }
    for(int i=1;i<=m;i++){
        for(int j=1;j<=m;j++){
            if(intersect(L[i],L[j])){
                uni(L[i].a,L[j].a);
                uni(L[i].b,L[j].b);
                uni(L[i].b,L[j].a);
                uni(L[i].a,L[j].b);
            }
        }
    }

    int tmp = fi(1);
    for(int i=1;i<=n;i++){
        if(fi(i)!=tmp){
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    return 0;
}