POJ 1127-Jack Straws（計算幾何 線段相交）

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 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. 

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.

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

CONNECTED 
NOT CONNECTED 
CONNECTED 
CONNECTED 
NOT CONNECTED 
CONNECTED
CONNECTED
CONNECTED
CONNECTED

East Central North America 1994

N根木棍，木棍i兩端的座標分別是(Pix,Piy)和(Qix,Qiy).

給出若干木棍(Ai,Bi)，判斷每對木棍是否相連。

當兩根木棍之間有公共點時，就認為它們是相連的。通過相連的木棍間接連在一起的兩根木棍也被認為是相連的。

木棍看成二維平面上的線段，求解交點再判斷。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<queue>
#include<malloc.h>
using namespace std;
typedef long long ll;
const int maxn = 10000;
const int INF = 0x3f3f3f3f;

double EPS=1e-10;
double add(double a,double b)//double加法運算考慮誤差
{
    if(abs(a+b)<EPS*(abs(a)+abs(b))) return 0;
    return a+b;
}
struct P//結構體定義二維向量
{
    double x,y;
    P() {}
    P(double x,double y):x(x),y(y) {}
    P operator +(P p)
    {
        return P(add(x,p.x),add(y,p.y));
    }
    P operator -(P p)
    {
        return P(add(x,-p.x),add(y,-p.y));
    }
    P operator *(double d)
    {
        return P(x*d,y*d);
    }
    double dot(P p)//內積
    {
        return add(x*p.x,y*p.y);
    }
    double det(P p)//外積
    {
        return add(x*p.y,-y*p.x);
    }
};
bool on_seg(P p1,P p2,P q)//判斷點q是否線上段p1-p2上
{
    return (p1-q).det(p2-q)==0&&(p1-q).dot(p2-q)<=0;
}
P intersection(P p1,P p2,P q1,P q2)//計算直線p1-p2與q1-q2的交點
{
    return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));
}
int n,m;
P p[maxn],q[maxn];
int a[maxn],b[maxn];
bool g[maxn][maxn];//相連關係圖
void solve()
{
    int i,j,k;
    for(i=0; i<n; ++i)
    {
        g[i][i]=true;
        for(j=0; j<i; ++j)//判斷木棍i和j是否有公共點
        {
            if((p[i]-q[i]).det(p[j]-q[j])==0)//平行
                g[i][j]=g[j][i]=on_seg(p[i],q[i],p[j])||on_seg(p[i],q[i],q[j])||on_seg(p[j],q[j],p[i])||on_seg(p[j],q[j],q[i]);
            else//相交
            {
                P 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(k=0; k<n; ++k)//Floyd-Warshall演算法判斷任意兩點間是否相連
        for(i=0; i<n; ++i)
            for(j=0; j<n; ++j)
                g[i][j] |= g[i][k] && g[k][j];
    for(i=0; i<m; ++i)
        puts(g[a[i]-1][b[i]-1]?"CONNECTED ":"NOT CONNECTED ");
}
int main()
{
    while(1)
    {
        scanf("%d",&n);
        if(n==0) break;
        m=0;
        for(int i=0; i<n; ++i)
            scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&q[i].x,&q[i].y);
        while(1)
        {
            int aa,bb;
            scanf("%d%d",&aa,&bb);
            if(aa==0&&bb==0)break;
            a[m]=aa;
            b[m++]=bb;
        }
        solve();
    }
    return 0;
}
/**
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
**/