POJ 1127 Jack Straws 几何基础

Description

In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try to remove them one-by-one without disturbing the other straws. Here, we are only concerned with if various pairs of straws are connected by a path of touching straws. You will be given a list of the endpoints for some straws (as if they were dumped on a large piece of graph paper) and then will be asked if various pairs of straws are connected. Note that touching is connecting, but also two straws can be connected indirectly via other connected straws.

Input

Input consist multiple case,each case consists of multiple lines. The first line will be an integer n (1 < n < 13) giving the number of straws on the table. Each of the next n lines contain 4 positive integers,x1,y1,x2 and y2, giving the coordinates, (x1,y1),(x2,y2) of the endpoints of a single straw. All coordinates will be less than 100. (Note that the straws will be of varying lengths.) The first straw entered will be known as straw #1, the second as straw #2, and so on. The remaining lines of the current case(except for the final line) will each contain two positive integers, a and b, both between 1 and n, inclusive. You are to determine if straw a can be connected to straw b. When a = 0 = b, the current case is terminated.

When n=0,the input is terminated.

There will be no illegal input and there are no zero-length straws.

Output

You should generate a line of output for each line containing a pair a and b, except the final line where a = 0 = b. The line should say simply "CONNECTED", if straw a is connected to straw b, or "NOT CONNECTED", if straw a is not connected to straw b. For our purposes, a straw is considered connected to itself.

Sample Input

7
1 6 3 3 
4 6 4 9 
4 5 6 7 
1 4 3 5 
3 5 5 5 
5 2 6 3 
5 4 7 2 
1 4 
1 6 
3 3 
6 7 
2 3 
1 3 
0 0

2
0 2 0 0
0 0 0 1
1 1
2 2
1 2
0 0

0

Sample Output

CONNECTED 
NOT CONNECTED 
CONNECTED 
CONNECTED 
NOT CONNECTED 
CONNECTED
CONNECTED
CONNECTED
CONNECTED

给定n个木棒，两端分别是(x1,y1),(x2,y2)，但1木棒与2木棒相交，2木棒与3木棒相交时，说明1木棒与3木棒也相交
有若干个询问a和b，求a与b是否相交。判断两个木棒是否相交，可以想判断他们是否平行，平行的话直接其中一个木棒的一点在另一木棒上就是相交，不平行的话，求两个直线的交点，看这个交点是否在这两个木棒上在的话就是相交了。
下面是代码，根据挑战程序设计上的计算几何基础写的。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cmath>
using namespace std;
const int N = 20;
const int M = 110;
double EPS = 1e-10;
double add(double a,double b) {
    if(abs(a+b) < EPS * (abs(a) + abs(b))) return 0;
    return a + b;
}
struct Point{
    double x, y;
    Point(){}
    Point(double x, double y) :x(x), y(y) {}
    Point operator + (Point p) {
        return Point(add(x, p.x), add(y, p.y));
    }
    Point operator - (Point p) {
        return Point(add(x, -p.x), add(y, -p.y));
    }
    Point operator * (double d) {
        return Point(x*d, y*d);
    }
    double dot(Point p) {    //内积
        return add(x * p.x, y * p.y);
    }
    double det(Point p) {    //外积
        return add(x * p.y, -y * p.x);
    }
};

bool on_seg(Point p1, Point p2, Point q) {      //判断点q是否在线段p1-p2上
    return (p1 - q).det(p2 - q) == 0 && (p1 - q).dot(p2 - q) <= 0;
}
Point intersection(Point p1, Point p2, Point q1, Point q2) {    //计算直线p1-p2与直接q1-q2的交点
    return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
}
int n;
Point p[N], q[N];
int m;
bool g[N][N];
void solve() {
    for(int i = 1; i <= n; i ++) {
        g[i][i] = true;
        for(int j = 1; j < i; j ++) {
            if((p[i] - q[i]).det(p[j] - q[j]) == 0) {
                g[i][j] = g[j][i] = on_seg(p[i], q[i], q[j])
                                    || on_seg(p[i], q[i], p[j])
                                    || on_seg(p[j], q[j], q[i])
                                    || on_seg(p[j], q[j], p[i]);
            } else {
                Point r = intersection(p[i], q[i], p[j], q[j]);
                g[i][j] = g[j][i] = on_seg(p[i], q[i], r) && on_seg(p[j], q[j], r);
            }
        }
    }
    for(int k = 1; k <= n; k ++) {
        for(int i = 1; i <= n; i ++) {
            for(int j = 1; j <= n; j ++) {
                g[i][j] |= g[i][k] && g[k][j];
            }
        }
    }
}
int main() {
    while(cin >> n && n) {
        memset(g, false, sizeof(g));
        for(int i = 1; i <= n; i ++) {
            cin >> p[i].x >> p[i].y >> q[i].x >> q[i].y;
        }
        int a, b;
        solve();
        while(cin >> a >> b && (a+b)) {
            if(g[a][b]) printf("CONNECTED
");
            else printf("NOT CONNECTED
");
        }
    }
    return 0;
}