bzoj1007 [HNOI2008]水平可见直线

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在xoy直角坐标平面上有n条直线L1,L2,...Ln,若在y值为正无穷大处往下看,能见到Li的某个子线段,则称Li为可见的,否则Li为被覆盖的.
例如,对于直线:
L1:y=x; L2:y=-x; L3:y=0
则L1和L2是可见的,L3是被覆盖的.

给出n条直线,表示成y=Ax+B的形式(|A|,|B|<=500000),且n条直线两两不重合.求出所有可见的直线.

似乎随便搞一下就过了？大概只是维护一下单调性吧我还在想会不会有什么反例之类的……结果就1A了

最近视力不好喜欢用大字体

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<cstdlib>#include<cstring>#define LL long long#define fo(i,a,b) for(int i=a;i<=b;i++)using namespace std;#define N 50005#define eps 1e-8inline int read(){        int f=1,d=0;char s=getchar();        while(s<'0'||s>'9'){if (s=='-')f=-1;s=getchar();}        while(s>='0'&&s<'9'){d=d*10+s-'0';s=getchar();}        return f*d;}struct xian{        double k,b;        int dfn;}a[N];int n;bool ans[N];bool com(xian a,xian b){        if (fabs(a.k-b.k)<=eps) return a.b<b.b;        else return a.k<b.k;}double calc(int x,int y){        return (double)(a[y].b-a[x].b)/(a[x].k-a[y].k);}int s[N],top=0;void ins(int x){        while(top)        {                if (fabs(a[s[top]].k-a[x].k)<=eps) top--;else                if (top>1&&(double)calc(s[top],s[top-1])>=calc(s[top],x)) top--;                else break;        }        s[++top]=x;}int main(){        memset(a,0,sizeof(a));        n=read();        fo(i,1,n)        {                scanf("%lf%lf",&a[i].k,&a[i].b);                a[i].dfn=i;        }        sort(a+1,a+n+1,com);        fo(i,1,n)ins(i);        fo(i,1,top)ans[a[s[i]].dfn]=1;        fo(i,1,n)        if (ans[i]) printf("%d ",i);        return 0;}

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