问题描述：
Problem 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 5
Sample Output
1
2
2
2
5
问题思路：
P命令插入一条线段。
Q命令查询线段K所在线段集合的线段数目。
套用了判断两条线段是否相交的模板。
代码：
#include<stdio.h>
#include<algorithm>
using namespace std;
int pre[1007];
struct point
{
double x;
double y;
};
double mult(point a,point b,point c)
{
return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
bool check(point aa,point bb,point cc,point dd)
{
if(max(aa.x,bb.x)<min(cc.x,dd.x))
return false;
if(max(aa.y,bb.y)<min(cc.y,dd.y))
return false;
if (max(cc.x,dd.x)<min(aa.x,bb.x))
return false;
if(max(cc.y,dd.y)<min(aa.y,bb.y))
return false;
if(mult(cc,bb,aa)*mult(bb,dd,aa)<0)
return false;
if(mult(aa,dd,cc)*mult(dd,bb,cc)<0)
return false;
return true;
}
int find(int x)
{
int r=x;
while(r!=pre[r])
r=pre[r];
int i=x,j;
while(i!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void mix(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
pre[fy]=fx;
}
void compress(int x)
{
int r=x;
while(r!=pre[r])
r=pre[r];
pre[x]=r;
}
int main()
{
int n,T,count_p;
char c;
struct point p1[1007];
struct point p2[1007];
while(scanf("%d",&T)!=EOF)
{
while(T--)
{
scanf("%d",&n);
getchar();
count_p=0;
for(int i=1;i<=n;i++)
pre[i]=i;
for(int i=1;i<=n;i++)
{
c=getchar();
if(c==‘P‘)
{
count_p++;
scanf("%lf%lf%lf%lf",&p1[count_p].x,&p1[count_p].y,&p2[count_p].x,&p2[count_p].y);
getchar();
for(int j=1;j<count_p;j++)
if(check(p1[count_p],p2[count_p],p1[j],p2[j]))
mix(count_p,j);
}
else{
int x;
int count=0;
scanf("%d",&x);
getchar();
for(int i=1;i<=n;i++)
compress(i);
int root=find(x);
for(int i=1;i<=n;i++)
if(pre[i]==root)
count++;
printf("%d\n",count);
}
}
if(T!=0)
printf("\n");
}
}
return 0;
}