Jack Straws POJ - 1127 （几何计算）

Jack Straws

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5428		Accepted: 2461

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

题意：问两条线段是否连通，通过第三条线段连通也算连通
题解：几何计算的模版加并查集，用floyd算法应该也可以吧

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
const  int maxn=;
const double eps=1e-;

//考虑误差的加法运算
double add(double x,double y)
{
    if(abs(x+y)<eps*(abs(x)+abs(y)))
        return ;
    return x+y;
}

//二维向量结构体
struct P
{
    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)     //内积
    {
        return add(x*p.x,y*p.y);
    }
    double det (P p)   //外积
    {
        return add(x*p.y,-y*p.x);
    }
};

//判断点是否在直线上
bool on_seg(P p1,P p2,P q)
{
    return (p1-q).det(p2-q)== && (p1-q).dot(p2-q)<=;
}

//计算直线p1-p2与直线q1-q2的交点
P inter(P p1,P p2,P q1,P q2)
{
    return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));
}

int n;
P p[maxn],q[maxn];    //保存一条线段的两个端点
bool G[maxn][maxn];    //线段之间是否联通的图

 int main()
{
    while(cin>>n && n)
    {
        memset(G,false,sizeof(G));
        for(int i=;i<n;i++)
            cin>>p[i].x>>p[i].y>>q[i].x>>q[i].y;

        for(int i=;i<n;i++)
            for(int j=;j<n;j++)
            {
                if((p[i]-q[i]).det(p[j]-q[j])==)
                {
                    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
                {
                    P r=inter(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=;k<n;k++)
            for(int i=;i<n;i++)
                for(int j=;j<n;j++)
                    G[i][j] |=G[i][k] && G[k][j];
        int x,y;
        while(cin>>x>>y && (x||y))
        {
            x--;
            y--;
            if(G[x][y])
                cout<<"CONNECTED"<<endl;
            else
                cout<<"NOT CONNECTED"<<endl;
        }
    }
    return ;
}