POJ 1127 Jack Straws (线段相交)

题意：给定一堆线段，然后有询问，问这两个线段是不是相交，并且如果间接相交也可以。

析：可以用并查集和线段相交来做，也可以用Floyd来做，相交就是一个模板题。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Point{
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) { }
};
typedef Point Vector;
int dcmp(double x){
  if(fabs(x) < eps) return 0;
  return x < 0 ? -1 : 1;
}
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); }
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; }

struct Node{
  Point p, q;
};
Node a[maxn];
int p[maxn];
int Find(int x){ return x == p[x] ? x : p[x] = Find(p[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 judge(int i, int j){
  if(onSegment(a[i].p, a[j].p, a[j].q))  return true;
  if(onSegment(a[i].q, a[j].p, a[j].q))  return true;
  if(onSegment(a[j].p, a[i].p, a[i].q))  return true;
  if(onSegment(a[j].q, a[i].p, a[i].q))  return true;
  if(a[i].p == a[j].q || a[i].p == a[j].p || a[i].q == a[j].p || a[i].q == a[j].q)  return true;
  double c1 = Cross(a[i].q-a[i].p, a[j].p-a[i].p);
  double c2 = Cross(a[i].q-a[i].p, a[j].q-a[i].p);
  double c3 = Cross(a[j].q-a[j].p, a[i].p-a[j].p);
  double c4 = Cross(a[j].q-a[j].p, a[i].q-a[j].p);
  return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

int main(){
  while(scanf("%d", &n) == 1 && n){
    for(int i = 1; i <= n; ++i){
      p[i] = i;
      scanf("%lf %lf %lf %lf", &a[i].p.x, &a[i].p.y, &a[i].q.x, &a[i].q.y);
    }
    for(int i = 1; i <= n; ++i){
      int x = Find(i);
      for(int j = 1 + i; j <= n; ++j){
        int y = Find(j);
        if(x == y)  continue;
        if(judge(i, j))  p[y] = x;
      }
    }
    int x, y;
    while(scanf("%d %d", &x, &y) == 2 && x+y){
      x = Find(x);
      y = Find(y);
      if(x == y)  puts("CONNECTED");
      else  puts("NOT CONNECTED");
    }
  }
  return 0;
}