hdu 1558 Segment set 线段相交+并查集

Segment set

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

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

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<bitset>
#include<set>
#include<map>
#include<time.h>
using namespace std;
#define LL long long
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e3+10,M=2e6+10,inf=1e9+10;
const LL INF=1e18+10,mod=998244353,MOD=998244353;
const double eps=1e-8,pi=(4*atan(1.0));

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;
    }
//绕原点旋转角度B（弧度值），后x,y的变化
    void transXY(double B)
    {
        double tx = x,ty = y;
        x= tx*cos(B) - ty*sin(B);
        y= tx*sin(B) + ty*cos(B);
    }
};
struct Line
{
    Point s,e;
    Line() {}
    Line(Point _s,Point _e)
    {
        s = _s;
        e = _e;
    }
//两直线相交求交点 //第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交 //只有第一个值为2时，交点才有意义
    pair<int,Point> operator &(const Line &b)const
    {
        Point res = s;
        if(sgn((s-e)^(b.s-b.e)) == 0)
        {
            if(sgn((s-b.e)^(b.s-b.e)) == 0) return make_pair(0,res);//重合
            else return make_pair(1,res);//平行
        }
        double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        res.x += (e.x-s.x)*t;
        res.y += (e.y-s.y)*t;
        return make_pair(2,res);
    }
};
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;
}
char ch[N];
int fa[N],si[N];
int Find(int x)
{
    return x==fa[x]?x:fa[x]=Find(fa[x]);
}
void update(int u,int v)
{
    int x=Find(u);
    int z=Find(v);
    if(x!=z)
    {
        fa[x]=z;
        si[z]+=si[x];
    }
}
Line a[N];
int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        if(cas)printf("
");
        cas++;
        int cnt=0;
        for(int i=1;i<=1000;i++)
            fa[i]=i,si[i]=1;
        int q;
        scanf("%d",&q);
        while(q--)
        {
            scanf("%s",ch);
            if(ch[0]=='Q')
            {
                int x;
                scanf("%d",&x);
                int f=Find(x);
                printf("%d
",si[f]);
            }
            else
            {
                double x1,y1,x2,y2;
                scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
                Point s(x1,y1),e(x2,y2);
                a[++cnt]=Line(s,e);
                for(int i=1;i<cnt;i++)
                    if(inter(a[i],a[cnt]))update(i,cnt);
            }
        }
    }
    return 0;
}