hdu 1558 Segment set (并查集)

Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3907    Accepted Submission(s): 1471

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

 

题目大意：在一个平面直角坐标系里面，通过P操作不断的增加线段，假设两个线段有相交。就表明他们是一个集合里面的。Q操作询问当前情况下第k条线段所在的集合里面有几条线段。

并查集的题目，可是我认为主要考几何。我開始能够想到。通过推断两条线段是否有交点，假设有就放在一个集合里面。这么想的确非常easy，可是做起来真的十分麻烦。。

假设对于两条线段，能够通过简单计算得到两者的交点x0=(b2-b1)/(k1-k2)，还有y0。

那么我仅仅要推断x0,y0是否在线段相交的地方就可以。可是还要注意，这个交点是从k1,k2得到的。所以假设k1,k2不存在，又要分情况讨论。

下面是我的代码，感觉好像还有遗漏的地方，尽管的确是AC了。

#include<stdio.h>
#include<string.h>
int p[10000],sum[10000];
double x1[1005],x2[1005],y1[1005],y2[1005];
void init(int x)
{
    int i;
    for(i=0;i<=x;i++)
    p[i]=i;
    for(i=0;i<=x;i++)
    sum[i]=1;
}
int findroot(int x)
{
    int r=x;
    while(r!=p[r])
    r=p[r];
    int i,j;
    i=x;
    while(i!=r)
    {
        j=p[i];
        p[i]=r;
        i=j;
    }
    return r;
}
void merge(int x,int y)
{
    int fx=findroot(x);
    int fy=findroot(y);
    if(fx!=fy){
        p[fx]=fy;
        sum[fy]+=sum[fx];
            }
}
double jiaodian(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)
{
    
   
    
    if(x1==x2&&x3!=x4){             //k1不存在,k2存在
         double k2=(y3-y4)/(x3-x4);
        double y=k2*(x1-x3)+y3;
        if((y>=y1&&y<=y2)||(y>=y2&&y<=y1))return 1;
        else return 0;
    }
   else if(x3==x4&&x1!=x2){           //k2不存在，k1存在
        double k1=(y1-y2)/(x1-x2);
        double y=k1*(x3-x1)+y1;
        if((y>=y3&&y<=y4)||(y>=y4&&y<=y3))return 1;
        else return 0;
    }
   else if(x1==x2&&x3==x4){
        if(x1==x3&&((y1>=y3&&y1<=y4)||(y1>=y4&&y1<=y3)||(y2>=y4&&y2<=y3)||(y2>=y3&&y2<=y4)))return 1;
        else return 0;
    }  
    double k1=(y1-y2)/(x1-x2);
     double k2=(y3-y4)/(x3-x4);
    double b1=(x1*y2-x2*y1)/(x1-x2);
    double b2=(x3*y4-x4*y3)/(x3-x4);
    double x=(b2-b1)/(k1-k2);
    double y=k1*(x-x1)+y1;
    if(((x>=x1&&x<=x2)||(x>=x2&&x<=x1))&&((y>=y1&&y<=y2)||(y>=y2&&y<=y1))&&
    ((x>=x3&&x<=x4)||(x>=x4&&x<=x3))||((y>=y3&&y<=y4)&&(y>=y4&&y<=y3)))return 1;
     return 0;
}
void isconnect(int x)
{
    int i;
    for(i=1;i<=x;i++)
  {
      
      if(jiaodian(x1[i],y1[i],x2[i],y2[i],x1[x],y1[x],x2[x],y2[x])){merge(i,x);}
  }
  return ;
}
int main()
{
    int t,n,i,j,k,m,cnt,q;
     
    char c[10];
    scanf("%d",&t);
    while(t--)
    {
        q=1;
        scanf("%d",&n);
        init(n);
        
        cnt=1;
        for(i=1;i<=n;i++)
        {
            scanf("%s",c);
            if(c[0]=='P')
            {
            scanf("%lf%lf%lf%lf",&x1[q],&y1[q],&x2[q],&y2[q]);
        
            if(i>1){
             isconnect(q);
                 
            }
            q++;
            }
            if(c[0]=='Q'){
                scanf("%d",&k);
                int s=findroot(k);
                cnt=sum[s];
                printf("%d
",cnt);
            
            }
            
        }
            if(t>0)printf("
");
    }
    return 0;
}