HDU 1558 计算几何+并查集

Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2896    Accepted Submission(s): 1073

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

线段相交判断+并查集。

代码：

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>

using namespace std;
const double eps = 1e-8;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x = _x;y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    //叉积
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    //点积
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;e = _e;
    }
}pp[1010];
bool inter(Line l1,Line l2)
{
    return
    max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
    max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
    max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
    max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
    sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&
    sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;
}
int fa[1010],num[1010];
int find(int x)
{
        if(fa[x]!=x)fa[x]=find(fa[x]);
        return fa[x];
}
void uni(int x,int y)
{
        if(x==y)return;
        if(inter(pp[x],pp[y]))
        {
                int fx=find(x);
                int fy=find(y);
                if(fx==fy)return;
                fa[fx]=fy;
             //   num[fx]+=num[fy];
                num[fy]+=num[fx];
        }

}
int n,m;
int main()
{
        int i,j,k,T;char ch[30];
        scanf("%d",&T);
        while(T--)
        {
               n=0;
               scanf("%d",&m);memset(num,0,sizeof(num));
               while(m--)
               {

                       scanf("%s",ch);
                       if(ch[0]=='P')
                       {
                            Point p,q;
                            scanf("%lf%lf%lf%lf",&p.x,&p.y,&q.x,&q.y);
                            pp[++n]=Line(p,q);
                            fa[n]=n;num[n]++;
                            for(i=1;i<n;i++)uni(i,n);
                       }
                       else
                       {
                               scanf("%d",&i);
                               int fx=find(i);
                               printf("%d\n",num[fx]);
                       }
               }
               if(T)puts("");
        }
        return  0;
}