Segment set

Segment set

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 231 Accepted Submission(s): 104

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

 

Author

LL

 

Source

HDU 2006-12 Programming Contest

 

Recommend

LL

/*
将有能通过某条线段连通的两条线段加入同一个并查集
*/
#include<bits/stdc++.h>
using namespace std;
//点的结构体
struct point
{
    double x,y;
    point(){}
    point(double a,double b)
    {
        x=a;
        y=b;
    }
};
struct line
{
    point a,b;
    line(){}
    line(point x,point y)
    {
        a=x;
        b=y;
    }
};
int t,n;
char op[2];
vector<line>v;
//叉积
double cross(const point &a,const point &b,const point &o)
{
    return (a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
}
//判断两条直线是否相交，端点重合算相交
bool IsIntersect(const point &a,const point &b,const point &c,const point &d)
{
    return ((cross(a,d,c)*cross(d,b,c)>=0)&&(cross(c,b,a)*cross(b,d,a)>=0)&&(max(c.x,d.x)>=min(a.x,b.x))&&(max(a.x,b.x)>=min(c.x,d.x))&&(max(c.y,d.y)>=min(a.y,b.y))&&(max(a.y,b.y)>=min(c.y,d.y)));
}
int bin[1010];
point x1,x2;
void inti()
{
    for(int i=0;i<=n;i++)
        bin[i]=i;
}
int findx(int x)
{
    while(x!=bin[x])
        x=bin[x];
    return x;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        if(ca!=1)
            printf("\n");
        //for(int i=0;i<=n;i++)
        //    cout<<bin[i]<<" ";
        //cout<<endl;
        scanf("%d",&n);
        v.clear();
        inti();
        while(n--)
        {
            scanf("%s",op);
            if(op[0]=='P')//添加边
            {
                scanf("%lf%lf%lf%lf",&x1.x,&x1.y,&x2.x,&x2.y);
                v.push_back(line(x1,x2));
                int cur=v.size();
                line temp=v[cur-1];
                for(int i=1;i<v.size();i++)
                {
                    if(IsIntersect(temp.a,temp.b,v[i-1].a,v[i-1].b))//两条线段有交点
                    {
                        //cout<<"i="<<i<<" cur="<<cur<<endl;
                        int fx=findx(i);
                        int fy=findx(cur);
                        //for(int i=0;i<=n;i++)
                        //    cout<<bin[i]<<" ";
                        //cout<<endl;
                        if(fx!=fy)
                            bin[fx]=fy;
                    }
                }
            }
            else//询问与那条边关联
            {
                int q;
                int cnt=0;
                scanf("%d",&q);
                for(int i=1;i<=v.size();i++)
                {
                    int fx=findx(i);
                    int fy=findx(q);
                    //cout<<"fx="<<fx<<" fy="<<fy<<endl;
                    if(fx==fy)
                    {
                        //cout<<"i="<<i<<" q="<<q<<endl;
                        cnt++;
                    }
                }
                printf("%d\n",cnt);
            }
        }
    }
    return 0;
}