PKU2036

对线段排序 对于K不存在的线段的放在一起,对x1进行排序 对于x1相同的对于y1排序
对于K存在的直线 对b进行排序 b相同的对于x1进行排序
排序复杂度Nlogn  在对于在一条直线上的线段O(n)的计算线段数目
总时间复杂度NLOGN这应该是这题最基本的算法

#include  < stdio.h >
#include  < string .h >
#include  < stdlib.h >
#include  < math.h >

struct  line
{
    double x1, y1, x2, y2;
    double b, k;
    bool ex_k;
} ;

line l[ 10000 ];
int  result;
int  cnt;

void  func( int  begin,  int  end)
{
    int i;
    int c=0;
    double back_x, back_y;
    for(i = begin; i <= end; i++)
    {
        if(l[i].ex_k)
        {
            if(i == begin || back_x < l[i].x1)
                c++;
            if(i == begin || back_x < l[i].x2)
                back_x=l[i].x2;
        }
        else
        {
            if(i == begin || back_y < l[i].y1)
                c++;
            if(i == begin || back_y < l[i].y2)
                back_y=l[i].y2;
        }
    }
    result += c;
}

bool  eq( const   double  a,  const   double  b)
{
    if(fabs(a-b)<0.00000001)
        return 1;
    else
        return 0;
}

int  cmp1( const   void   *  a,  const   void   *  b)
{
    line * aa = (line *)a;
    line * bb = (line *)b;
    if(aa->ex_k != bb->ex_k )
        return aa->ex_k > bb->ex_k ? 1 : -1;
    else 
        if(aa->ex_k && bb->ex_k)
        {
            if(!eq(aa->k , bb->k))
                return aa->k > bb->k ? 1 : -1;
            else
                if(!eq(aa->b , bb->b))
                    return aa->b > bb->b ? 1 : -1;
                else
                    return aa->x1 > bb->x1 ? 1 : -1;
        }
        else
            if(!eq(aa->x1 , bb->x1))
                return aa->x1 > bb->x1 ? 1 : -1;
            else
                return aa->y1 > bb->y1 ? 1 : -1;
}

int  main()
{
    int n;
    while(scanf("%d", &n) == 1 && n)
    {
        result = 0;
        cnt = 0;
        memset(l,0,sizeof(l));
        int i;
        line ll;
        for(i = 0; i < n; i++)
        {
            scanf("%lf%lf%lf%lf", &l[i].x1, &l[i].y1, &l[i].x2, &l[i].y2);
            if(l[i].x1==l[i].x2)
            {
                l[i].ex_k=0;
                if(l[i].y1 > l[i].y2)
                {
                    double temp=l[i].x1;
                    l[i].x1=l[i].x2;
                    l[i].x2=temp;
                    temp=l[i].y1;
                    l[i].y1=l[i].y2;
                    l[i].y2=temp;
                }
            }
            else
            {
                l[i].ex_k=1;
                if(l[i].x1 > l[i].x2)
                {
                    double temp=l[i].x1;
                    l[i].x1=l[i].x2;
                    l[i].x2=temp;
                    temp=l[i].y1;
                    l[i].y1=l[i].y2;
                    l[i].y2=temp;
                }
                l[i].k=(l[i].y2-l[i].y1)/(l[i].x2-l[i].x1);
                l[i].b=l[i].y1-l[i].x1*(l[i].y2-l[i].y1)/(l[i].x2-l[i].x1);            
            }
        }

        qsort(l, n , sizeof(l[0]), cmp1);

        int begin = 0, end = 0;
        for(i = 0; i < n; i++)
        {
            if(i == n-1 || l[i].ex_k != l[i+1].ex_k || !l[i].ex_k&&!eq(l[i].x1 , l[i+1].x1) || !eq(l[i].k , l[i+1].k) || !eq(l[i].b , l[i+1].b) )
            {
                end = i;
                func(begin, end);
                begin = i+1;
            }
        }
        printf("%d\n", result);

    }
    return 0;
}