poj1127 Jack Straws 线段相交+并查集

最新推荐文章于 2020-03-05 13:20:14 发布

原创最新推荐文章于 2020-03-05 13:20:14 发布 · 1k 阅读

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博客探讨了如何使用线段相交算法和并查集数据结构来解决POJ1127问题，即判断平面上的n条线段之间是否存在直接或间接的连接关系。文章补充了线段相交模板中对共线但不相交情况的处理方法。

平面上有n条线段，询问某两条线段是否直接或间接相连。

才发现以前写的线段相交模板中没有处理共线但不相交的情况，现在加上了。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <cstdio>
#include <algorithm>
#define N 10005
using namespace std;
typedef long long LL;
struct point
{
  LL x , y;
  point (LL x = 0 , LL y = 0): x(x) , y(y) {};
  void input() {cin >> x >> y;}
  bool operator < (const point& r) const{
    if (x == r.x)
      return y < r.y;
    return x < r.x;}
  bool operator == (const point& r) const{
    return x == r.x && y == r.y;
    }
};
point operator + (point a , point b) { return point(a.x + b.x , a.y + b.y);}
point operator - (point a , point b) { return point(a.x - b.x , a.y - b.y);}
LL Cross(point a , point b) {return a.x * b.y - a.y * b.x;}
double length(point a) {return sqrt(double(a.x) * a.x + double(a.y) * a.y);}
double dis(point a , point b) {return length(a - b);}
int dcmp(LL x) {return x == 0 ? 0 : x > 0 ? 1 : -1;}
bool SegmentI(point a1 , point a2 , point b1 , point b2)
{
  LL c1 = Cross(a2 - a1 , b1 - a1) , c2 = Cross(a2 - a1 , b2 - a1),
     c3 = Cross(b2 - b1 , a1 - b1) , c4 = Cross(b2 - b1 , a2 - b1);
  if (!c1 && !c2 && !c3 && !c4)
  {
    point x = a2 < b2 ? a2 : b2;
    point y = a1 < b1 ? b1 : a1;
    return x == y || y < x;
  }
  return dcmp(c1) * dcmp(c2) <= 0 && dcmp(c3) * dcmp(c4) <= 0;
}
int n , f[N];
point a[N] , b[N];
int getf(int x) {return x == f[x] ? x : f[x] = getf(f[x]);}
void work()
{
  int i , j;
  for (i = 1 ; i <= n ; ++ i){
    a[i].input() , b[i].input() , f[i] = i;
    if (b[i] < a[i]) swap(a[i] , b[i]);
  }
  for (i = 1 ; i <= n ; ++ i)
    for (j = i + 1 ; j <= n ; ++ j)
      if (SegmentI(a[i] , b[i] , a[j] , b[j]))
        f[getf(i)] = getf(j);
  while (scanf("%d%d",&i,&j) , i || j)
    puts(getf(i) == getf(j) ? "CONNECTED" : "NOT CONNECTED");
}


int main()
{
  while(scanf("%d",&n) , n)
  work();
  return 0;
}