[POJ2318]TOYS（计算几何+二分）

来源：互联网发布：前端程序员是干什么的编辑：程序博客网时间：2024/04/20 03:07

题目描述

传送门
题意：一些直线把一个矩形分割成了若干个部分。给出一些点问每一个部分里有多少个点

题解

叉积的性质：a×b，如果b在a的左边，叉积为正，右边为负，ab共线即为0
可以二分出所在部分的右边界，是否在某一条直线的右边可以用叉积判断一下

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;    }};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 ans[N];void clear(){    memset(ans,0,sizeof(ans));}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;        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));        for (int i=1;i<=m;++i)        {            scanf("%lf%lf",&x,&y);            int loc=find(Point(x,y));            ++ans[loc];        }        for (int i=0;i<=n;++i)            printf("%d: %d\n",i,ans[i+1]);        puts("");    }}

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