hdu 1558

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

这题重点是几何，还有就是注意输出的格式问题
代码

#include <iostream>

using namespace std;
int pre[1100];
int num[1100];
struct point
{
double x,y;
};
struct xian
{
point a,b;
}hehe[1100];
int find(int x)
{
int r=x;
while(pre[r]!=r)
{
r=pre[r];
}
int i=x,j;
while(i!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void join(int x,int y)
{
int fx=find(x),fy=find(y);
if(fx!=fy)
{
pre[fx]=fy;
num[fy]=num[fy]+num[fx];
}
}
double xm(point a,point b,point c)
{
return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
bool seg(point a,point b,point c)
{
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)
{
double d1,d2,d3,d4;
d1=xm(c,d,a);
d2=xm(c,d,b);
d3=xm(a,b,c);
d4=xm(a,b,d);
if(d1*d2<0&&d3*d4<0) return 1;
else if(d1==0&&seg(c,d,a)) return 1;
else if(d2==0&&seg(c,d,b)) return  1;
else if(d3==0&&seg(a,b,c)) return 1;
else if(d4==0&&seg(a,b,d)) return 1;
return 0;
}
int main()
{
int t,n,k,tmp;
char c;
cin>>t;
while(t--)
{
k=0;
cin>>n;
for(int i=1;i<=n;i++)
{
pre[i]=i;
num[i]=1;
}
for(int i=1;i<=n;i++)
{
cin>>c;
if(c=='P')
{
k++;
cin>>hehe[k].a.x>>hehe[k].a.y>>hehe[k].b.x>>hehe[k].b.y;
for(int j=1;j<k;j++)
{
if(cross(hehe[k].a,hehe[k].b,hehe[j].a,hehe[j].b))
join(j,k);
}
}
else
{
cin>>tmp;
cout<<num[find(tmp)]<<endl;
}
}
if(t)
cout<<endl;
}
return 0;
}