[POJ2398]Toy Storage（计算几何+二分）

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题目描述

传送门
题解

同POJ2318
代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<ctime>using namespace std;#define N 10005const double eps=1e-9;int dcmp(double x){    if (x<=eps&&x>=-eps) return 0;    return (x>0)?1:-1;}struct Vector{    double x,y;    Vector(double X=0,double Y=0)    {        x=X,y=Y;    }};typedef Vector Point;struct Line{    Point p;    Vector v;    Line(Point P=Point(0,0),Vector V=Vector(0,0))    {        p=P,v=V;    }    bool operator < (Line A) const    {        return p.x<A.p.x;    }};Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x,A.y-B.y);}int n,m;double a,b,c,d,ui,li,x,y;Line l[N];int cnt[N],ans[N];void clear(){    memset(cnt,0,sizeof(cnt));    memset(ans,0,sizeof(ans));}int cmp(Line a,Line b){    return a<b;}double Cross(Vector A,Vector B){    return A.x*B.y-A.y*B.x;}bool check(int mid,Point pt){    Point st=l[mid].p;    Vector c=Vector(pt.x-st.x,pt.y-st.y);    double t=Cross(l[mid].v,c);    if (dcmp(t)>0) return true;    else return false;}int find(Point pt){    int l=1,r=n+1,mid,ans=0;    while (l<=r)    {        mid=(l+r)>>1;        if (check(mid,pt)) ans=mid,r=mid-1;        else l=mid+1;    }    return ans;}int main(){    while (~scanf("%d",&n))    {        if (!n) break;        puts("Box");        clear();        scanf("%d",&m);        scanf("%lf%lf%lf%lf",&a,&b,&c,&d);        for (int i=1;i<=n;++i)        {            scanf("%lf%lf",&ui,&li);            l[i]=Line(Point(li,d),Vector(ui-li,b-d));        }        l[n+1]=Line(Point(c,d),Vector(0,b-d));        sort(l+1,l+n+2);        for (int i=1;i<=m;++i)        {            scanf("%lf%lf",&x,&y);            int loc=find(Point(x,y));            ++cnt[loc];        }        for (int i=1;i<=n+1;++i) ++ans[cnt[i]];        for (int i=1;i<=m;++i)            if (ans[i])                printf("%d: %d\n",i,ans[i]);    }}
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