POJ 1127 Jack Straws (线段相交 Floyd)

题目链接：http://poj.org/problem?id=1127

题意：给出n条线段，判断两条线段是否相交（此处的相交也满足传递性，即若a与b相交并且b与c相交，那么a与c也相交）

思路：对于相交的传递性，用floyd算法求解即可。线段相交的代码摘自入门经典。

#include<cstdio>
#include<cstring>
#include<string>
#include<cctype>
#include<iostream>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 13 + 5;
struct Point {
  double x, y;
  Point(double x=0, double y=0):x(x),y(y) {}
}a[maxn], b[maxn];
typedef Point Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x-B.x, A.y-B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

const double eps = 1e-10;
int dcmp(double x) {
  if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b) {
  return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }

bool OnSegment(Point p, Point a1, Point a2) {
  return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) <= 0;
}

bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
  double 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(dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0) return true;
  if(OnSegment(a1, b1, b2) || OnSegment(a2, b1, b2) || OnSegment(b1, a1, a2) || OnSegment(b2, a1, a2)) return true;
  return false;
}

int main() {
  bool dp[15][15]; int n;
  while(scanf("%d", &n) && n) {
    memset(dp, false, sizeof dp);
    for(int i = 1; i <= n; i++) scanf("%lf%lf%lf%lf", &a[i].x, &a[i].y, &b[i].x, &b[i].y);
    for(int i = 1; i <= n; i++)
      for(int j = i; j <= n; j++)
        if(i == j || SegmentProperIntersection(a[i], b[i], a[j], b[j]))
          dp[i][j] = dp[j][i] = true;
    for(int k = 1; k <= n; k++)
      for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
          dp[i][j] |= dp[i][k] & dp[k][j];
    int a, b;
    while(scanf("%d%d", &a, &b) && a+b) {
      if(dp[a][b] || dp[b][a]) printf("CONNECTED\n");
      else printf("NOT CONNECTED\n");
    }
  }
  return 0;
}