Jack Straws(并差集和判断直线相交问题)

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

意解: 并差集加计算几何知识 (注意判断直线共线但不相交的情况)
可以作为判断直线相交的模板...
AC代码:

#include <iostream>
#include <cstdio>

using namespace std;
int set[20];

struct Point
{
    int x1,x2,y1,y2;
    Point(int x1 = 0, int x2 = 0, int y1 = 0, int y2 = 0) : x1(x1),x2(x2),y1(y1),y2(y2) {};
    void read()
    {
        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
    }
}p[20];

void unit(int n)
{
    for(int i = 1; i <= n; i++) set[i] = i;
}

int find(int x)
{
    return x == set[x] ? x : set[x] = find(set[x]);
}

int cross(int x1, int y1, int x2, int y2)
{
    return x1 * y2 - x2 * y1;
}

int intersection(Point A, Point B) //判断直线相交
{
    int c[4];
    if(max(A.x1,A.x2) < min(B.x1,B.x2) || max(A.y1,A.y2) < min(B.y1,B.y2)
       || max(B.x1,B.x2) < min(A.x1,A.x2) || max(B.y1,B.y2) < min(A.y1,A.y2) ) return 0; //考虑共线不相交的情况,为快速排斥定理
    /*判断两条直线是否相交,即只需判断线是否在另一条线的两端*/
    c[0] = cross(A.x2 - A.x1, A.y2 - A.y1, B.x1 - A.x1, B.y1 - A.y1);
    c[1] = cross(A.x2 - A.x1, A.y2 - A.y1, B.x2 - A.x1, B.y2 - A.y1);
    c[2] = cross(B.x2 - B.x1, B.y2 - B.y1, A.x1 - B.x1, A.y1 - B.y1);
    c[3] = cross(B.x2 - B.x1, B.y2 - B.y1, A.x2 - B.x1, A.y2 - B.y1);
    if(c[0] * c[1] <= 0 && c[2] * c[3] <= 0) return 1; //运用到了向量的叉乘和点乘的知识;
    return 0;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while(~scanf("%d",&n) && n)
    {
        for(int i = 1; i <= n; i++)
            p[i].read();
        unit(n);
        for(int i = 1; i <= n; i++)
        {
            for(int j = i + 1; j <= n; j++)
            {
                if(intersection(p[i],p[j]))
                {
                    int a = find(i);
                    int b = find(j);
                    if(a != b) set[a] = b;
                }
            }
        }
        int a,b;
        while(~scanf("%d %d",&a,&b), a | b)
        {
            a = find(a);
            b = find(b);
            if(a == b) puts("CONNECTED");
            else puts("NOT CONNECTED");
        }
    }
    return 0;
}