Jack Straws
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 3945 | Accepted: 1787 |
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：
桌子上放着n根木棍，木棍的两端坐标分别是(Pix, Piy)和(Qix, Qiy)。给定m对木棍(ai, bi),请判断没对木棍是否相连。当两根木棍之间有公共点时，就认为它们是相连的。通过相连的木棍间接的连在一起的两根木棍也认为是相连的。
分析：
木棍就是二维平面上的线段，只有能够判断线段是否相交，那么建图以后就可以轻松的进行连接性判断。那么，应该如何判断两条线段是否相交呢？首先会想到计算两直线的交点，然后判断交点是否在线段上这一方法。那么两条直线的交点要怎么求得呢？虽然可以把直线表示成方程，通过联立方程组求解，但是在几何问题中，运用向量的内积和外积进行计算是非常方便的。对于二维向量p1=(x1, y1) 和 p2=(x2, y2)，我们定义内积p1*p2 = x1x2 +y1y2，外积p1*p2 = x1y2 – x2y1。要判断点q是否在线段p1-p2上，只有先利用外积根据是否有(p1-q)*(p2
– q)=0来判断点q是否在直线p1-p2上，在利用内积根据是否有(p1-q)*(p2– q)<=0来判断点q是否落在p1-p2之间。而要求两直线的交点，通过变量t将直线p1-p2上的点表示为p1+t(p2-p1)，交点又在直线q1-q2上，所以有：
(q2-q1)*(p1+t(p2-p1)-q1) = 0
于是可以利用下式求得t的值
P1+(q2-q1)*(q1-p1)*(p2-p1)/(q2-q1)*(p2-p1)
但是，使用这个方法时还有注意边界情况。让我们来看看样例中的木棍2和木棍4，这两条线段是平行的，对应直线没有交点。但平行的线段也可能有公共点，所有此时需要特别注意。对此有不同的处理方法，这里我们选择通过检查端点是否在另一条线段上来判断。
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
double EPS = 1e-10;
//考虑误差的加法运算
double add(double a, double b)
{
if (abs(a + b) < EPS * (abs(a) + abs(b)))
return 0;
return a + b;
}
//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dianq是否在线段p1-p2上
bool on_seg(P p1, P p2, P q)
{
return (p1 - q).det(p2 - q) == 0 && (p1 - q).dot(p2 - q) <= 0;
}
//计算直线p1-p2与直线q1-q2的交点
P intersection(P p1, P p2, P q1, P q2)
{
return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
}
//输入
int n;
P p[15], q[15];
int m;
int a, b;
bool g[15][15]; //相连关系图
void solve()
{
memset(g, false, sizeof(g));
for (int i = 0; i < n; i++){
g[i][i] = true;
for (int j = 0; j < i; j++){
//判断木棍i和木棍j是否有公共点
if ((p[i] - q[i]).det(p[j] - q[j]) == 0){
//平行时
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{
//非平行时
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);
}
}
}
//通过Floyd-Warshall算法判断任意两点间是否相连
for (int k = 0; k < n; k++){
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
g[i][j] |= g[i][k] && g[k][j];
}
}
}
while (scanf("%d%d", &a, &b) != EOF){
if (a == 0 && b == 0){
break;
}
puts(g[a - 1][b - 1] ? "CONNECTED" : "NOT CONNECTED");
}
}
int main()
{
while (scanf("%d", &n), n){
for (int i = 0; i < n; i++){
scanf("%lf%lf%lf%lf", &p[i].x, &p[i].y, &q[i].x, &q[i].y);
}
solve();
}
return 0;
}