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用到的计算几何中的内容至今不大懂，目前练习并查集，以后待续、、、、
1、题目大意：
给你一些操作，P后边输入四个值，分别代表一条线段的起点、终点坐标，
当输入Q时，后边输入一个整形值K，输出第k条线段所在的集合中包含的线段的个数
2、思路：用并查集做
当输入P时，判断后边输入的线段的起点和终点时，判断跟之前的线段有没有相交，如果有相交，就merge（）合并，
如果输入的是Q时，就打印出当前线段所在集合的个数
3、题目：
Segment set
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2386 Accepted Submission(s): 924Problem Description
A segment and all segments which are connected with it compose a segment set. The size of a segment set is the number of segments in it. The problem is to find the size of some segment set.Input
In the first line there is an integer t - the number of test case. For each test case in first line there is an integer n (n<=1000) - the number of commands. There are two different commands described in different format shown below:P x1 y1 x2 y2 - paint a segment whose coordinates of the two endpoints are (x1,y1),(x2,y2).
Q k - query the size of the segment set which contains the k-th segment.k is between 1 and the number of segments in the moment. There is no segment in the plane at first, so the first command is always a P-command.Output
For each Q-command, output the answer. There is a blank line between test cases.Sample Input
1
10
P 1.00 1.00 4.00 2.00
P 1.00 -2.00 8.00 4.00
Q 1
P 2.00 3.00 3.00 1.00
Q 1
Q 3
P 1.00 4.00 8.00 2.00
Q 2
P 3.00 3.00 6.00 -2.00
Q 5Sample Output
1
2
2
2
5Author
LLSource
HDU 2006-12 Programming Contest Recommend3、代码：
#include<stdio.h>
#include<iostream>
using namespace std;
int set[1010];
int num[1010];
struct point
{double x,y;
};
struct edge
{point a;point b;
}e[1010];
int find(int x)
{int r=x;while(r!=set[r])r=set[r];int i=x;while(i!=r){int j=set[i];set[i]=r;i=j;}return r;
}void merge(int x,int y)
{int fx=find(x);int fy=find(y);if(fx!=fy){set[fx]=fy;num[fy]+=num[fx];}
}
double xmult(point a,point b,point c) //大于零代表a,b,c左转
{return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
bool OnSegment(point a,point b,point c) //a,b,c共线时有效
{return c.x>=min(a.x,b.x)&&c.x<=max(a.x,b.x)&&c.y>=min(a.y,b.y)&&c.y<=max(a.y,b.y);
}bool Cross(point a,point b,point c,point d) //判断ab 与cd是否相交
{double d1,d2,d3,d4;d1=xmult(c,d,a);d2=xmult(c,d,b);d3=xmult(a,b,c);d4=xmult(a,b,d);if(d1*d2<0&&d3*d4<0) return 1;else if(d1==0&&OnSegment(c,d,a)) return 1;else if(d2==0&&OnSegment(c,d,b)) return 1;else if(d3==0&&OnSegment(a,b,c)) return 1;else if(d4==0&&OnSegment(a,b,d)) return 1;return 0;
}
int main()
{int t,n,k,tmp,i,j;char s[10];scanf("%d",&t);while(t--){scanf("%d",&n);k=0;for(int i=1;i<=n;i++){set[i]=i;num[i]=1;}for(int i=1;i<=n;i++){scanf("%s",s);if(s[0]=='P'){k++;scanf("%lf%lf%lf%lf",&e[k].a.x,&e[k].a.y,&e[k].b.x,&e[k].b.y);for(int j=1;j<k;j++){if(find(k)!=find(j)&&Cross(e[k].a,e[k].b,e[j].a,e[j].b))merge(k,j);}}else if(s[0]=='Q'){scanf("%d",&tmp);printf("%d\n",num[find(tmp)]);}}if(t)//There is a blank line between test cases.注意最后一个样例不用输出空行printf("\n");}return 0;
}
/*
2
10
P 1.00 1.00 4.00 2.00
P 1.00 -2.00 8.00 4.00
Q 1
P 2.00 3.00 3.00 1.00
Q 1
Q 3
P 1.00 4.00 8.00 2.00
Q 2
P 3.00 3.00 6.00 -2.00
Q 5
3
P 1.00 1.00 4.00 2.00
P 1.00 -2.00 8.00 4.00
Q 1
*/