BZOJ 1007 [HNOI2008]水平可见直线

题解：

维护一个栈就好了，画画图就知道怎么维护了。。

 

这两天总写不对计算几何的水题。。真是郁闷。。

 

View Code

  1 #include <iostream>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <cstdio>

  5 #include <algorithm>

  6 

  7 #define N 555555

  8 #define EPS 1e-7

  9 

 10 using namespace std;

 11 

 12 struct PO

 13 {

 14     double x,y;

 15     void prt() {printf("%lf       %lf\n",x,y);}

 16 }sk[N];

 17 

 18 struct LI

 19 {

 20     PO a,b;

 21     int id;

 22     double p,q;

 23     void prt() {printf("%lf     %lf     %lf     %lf\n",a.x,a.y,b.x,b.y);}

 24 }li[N],stk[N];

 25 

 26 int n,top,tp,bh[N];

 27 

 28 inline PO operator +(PO a,PO b)

 29 {

 30     a.x+=b.x; a.y+=b.y;

 31     return a;

 32 }

 33 

 34 inline PO operator -(PO a,PO b)

 35 {

 36     a.x-=b.x; a.y-=b.y;

 37     return a;

 38 }

 39 

 40 inline PO operator *(PO a,double k)

 41 {

 42     a.x*=k; a.y*=k;

 43     return a;

 44 }

 45 

 46 inline PO operator /(PO a,double k)

 47 {

 48     a.x/=k; a.y/=k;

 49     return a;

 50 }

 51 

 52 inline int dc(double x)

 53 {

 54     if(x>EPS) return 1;

 55     else if(x<-EPS) return -1;

 56     return 0;

 57 }

 58 

 59 inline void read()

 60 {

 61     scanf("%d",&n);

 62     double a,b;

 63     for(int i=1;i<=n;i++)

 64     {

 65         scanf("%lf%lf",&a,&b);

 66         li[i].a.x=0; li[i].a.y=b;

 67         if(dc(a)==0) li[i].b.x=29.3278,li[i].b.y=b;

 68         else li[i].b.x=29.3278,li[i].b.y=a*29.3278+b;

 69         li[i].p=a; li[i].q=b;

 70         li[i].id=i;

 71     }

 72 }

 73 

 74 inline double cross(const PO &a,const PO &b,const PO &c)

 75 {

 76     return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);

 77 }

 78 

 79 inline bool cmp(const LI &a,const LI &b)

 80 {

 81     if(dc(a.p-b.p)==0) return a.q>b.q;

 82     return a.p<b.p;

 83 }

 84 

 85 inline PO getpoint(const PO &a,const PO &b,const PO &c,const PO &d)

 86 {

 87     double k1=cross(a,d,c),k2=cross(b,c,d);

 88     PO tmp;tmp=b-a;

 89     return a+tmp*k1/(k1+k2);

 90 }

 91 

 92 inline void go()

 93 {

 94     PO c;

 95     sort(li+1,li+1+n,cmp);

 96     top=tp=0;

 97     for(int i=1;i<=n;i++)

 98     {

 99         if(dc(li[i].p-li[i-1].p)==0) continue;

100         while(tp>=1)

101         {

102             c=getpoint(li[i].a,li[i].b,stk[top].a,stk[top].b);

103             if(dc(c.x-sk[tp].x)<=0) top--,tp--;

104             else break;

105         }

106         stk[++top]=li[i];

107         if(top==2&&tp==0) sk[++tp]=getpoint(stk[top].a,stk[top].b,stk[top-1].a,stk[top-1].b);

108         else if(tp>0) sk[++tp]=c;

109     }

110     for(int i=1;i<=top;i++) bh[i]=stk[i].id;

111     sort(bh+1,bh+1+top);

112     for(int i=1;i<top;i++) printf("%d ",bh[i]);

113     printf("%d",bh[top]);

114 }

115 

116 int main()

117 {

118     read(),go();

119     return 0;

120 }