hdu1558——Segment set

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

判断和某一条线段相交的线段有几条，基础并查集，关键是判断线段相交不要弄错

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<iostream>

using namespace std;

struct node
{
	double x1,x2,y1,y2;
}line[1010];

int father[1010];
int rank[1010];
int find(int x)
{
	if(x==father[x])
	  return father[x];
    else
      father[x]=find(father[x]);
    return father[x];
}

void merge(int a,int b)
{
	int aa=find(a);
	int bb=find(b);
	if(aa!=bb)
	{
		father[aa]=bb;
		rank[bb]+=rank[aa];
	} 
}

bool judge(int a,int b)
{
	double min_ay=min(line[a].y1,line[a].y2);
	double max_ay=max(line[a].y1,line[a].y2);
	double min_by=min(line[b].y1,line[b].y2);
	double max_by=max(line[b].y1,line[b].y2);
	if(line[a].x1 == line[a].x2)
	{
		if(line[b].x1 == line[b].x2)
		{
			if(line[b].x1 != line[a].x1)//平行但不重合
			  return false;
			else
			{
				if(min_ay > line[b].y1 && min_ay > line[b].y2)  
				  return false;
		        else if(max_ay < line[b].y1 && max_ay < line[b].y2)
		          return false;
                else
                  return true;
			} 
		}
		else
		{
			double k2=(line[b].y1 - line[b].y2)/(line[b].x1 - line[b].x2);
			double b2=line[b].y1 - k2 * line[b].x1;
			double ly=k2 * line[a].x1 + b2;
			if(ly>=min_ay && ly <= max_ay && ly>=min_by && ly<=max_by)
			  return true;
	        else
	          return false;
		}
	}
	else
	{
		if(line[b].x1 == line[b].x2)
		{
			double k1=(line[a].y1 - line[a].y2)/(line[a].x1 - line[a].x2);
			double b1=line[a].y1 - k1 * line[a].x1;
			double ly=k1 * line[b].x1 + b1;
			if(ly>=min_ay && ly <= max_ay && ly>=min_by && ly<=max_by)
			  return true;
	        else
	          return false;
		}
		else//两条直线都有斜率 
		{
			double k1=(line[a].y1 - line[a].y2)/(line[a].x1 - line[a].x2);
			double b1=line[a].y1 - k1 * line[a].x1;
			double k2=(line[b].y1 - line[b].y2)/(line[b].x1 - line[b].x2);
			double b2=line[b].y1 - k2 * line[b].x1;
			if(k1 == k2)
			{
				if(b1 == b2)
				{
					if(min_ay > line[b].y1 && min_ay > line[b].y2)  
				      return false;
		            else if(max_ay < line[b].y1 && max_ay < line[b].y2)
		              return false;
                    else
                      return true;
				} 
				else
				    return false;
			}
			else
			{
				double ly=((b2-b1)/(k1-k2))*k1+b1;
				if(ly>=min_ay && ly <= max_ay && ly>=min_by && ly<=max_by)
			      return true;
	            else
	             return false;
			}
		}
	}
}


int main()
{
	int t;
	scanf("%d",&t);
	bool flag=false;
	while(t--)
	{
		if(flag)
			printf("\n");
		else
			flag=true;
		int n;
		scanf("%d",&n);
		for(int i=0;i<=n;i++)
		{
			rank[i]=1;
			father[i]=i;
		}
		char c[3];
		int cnt=1,res;
		double x1,y1,x2,y2;
		for(int i=0;i<n;i++)
		{
			scanf("%s",c);
			if(c[0]=='P')
			{
				scanf("%lf%lf%lf%lf",&line[cnt].x1,&line[cnt].y1,&line[cnt].x2,&line[cnt].y2);
				for(int i=1;i<cnt;i++)
				{
					if(judge(i,cnt))
						merge(i,cnt);
				}
				cnt++;
			}
			else
			{
				scanf("%d",&res);getchar();
				printf("%d\n",rank[find(res)]);
			}
		}
	}
	return 0;
}