hdu 1558 线段相交+并查集路径压缩

Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3457    Accepted Submission(s): 1290

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

 

题目大意：有n个指令，p加入一条线段,q查询id线段所在集合（两线段有交点为同一集合）的元素个数。

思路：用并查集路径压缩记录各个线段间的关系，根据叉积的定义有：Cross(v,w)=0时w在v上,>0时w在v上方,<0时w在v下方。

两线段有交点的必要条件：必须每条线段的两个端点在另一线段的两侧或直线上。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

const double eps=1e-8;
const int maxn=1005;
int f[maxn];
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
};
struct Line
{
    Point a,b;
}L[maxn];
typedef Point Vector;
Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
double Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x;}//叉积

bool judge(Line a,Line b)//Cross(v,w)=0时w在v上,>0时w在v上方,<0时w在v下方
{
    if(dcmp(Cross(a.a-b.a,b.b-b.a)*Cross(a.b-b.a,b.b-b.a))<=0
        &&dcmp(Cross(b.a-a.a,a.b-a.a)*Cross(b.b-a.a,a.b-a.a))<=0)
        return true;
    return false;
}
int findset(int x){return f[x]!=x?f[x]=findset(f[x]):x;}
void Union(int a,int b)
{
    a=findset(a);b=findset(b);
    if(a!=b) f[a]=b;
}
int main()
{
    int t,n,i,j,id;
    char op[5];
    double x1,y1,x2,y2;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int cnt=0;
        for(i=0;i<n;i++)
        {
            scanf("%s",op);
            if(op[0]=='P')
            {
                scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
                L[++cnt].a=Point(x1,y1);L[cnt].b=Point(x2,y2);
                f[cnt]=cnt;
                for(j=1;j<=cnt-1;j++)
                    if(judge(L[cnt],L[j]))
                        Union(j,cnt);
            }
            else 
            {
                int ans=0;scanf("%d",&id);
                id=findset(id);
                for(j=1;j<=cnt;j++)
                    if(findset(j)==id)
                        ans++;
                printf("%d
",ans);
            }
        }
        if(t) printf("
");
    }
    return 0;
}