POJ 2318 TOYS

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#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)#define oo 1<<30#define eps 1e-6#define nMax 6010int dcmp(double x){    if(fabs(x)<=eps) return 0;    else return x > 0?1:-1;}struct Point{    double x,y;    Point (double x=0,double y=0) : x(x),y(y) {}    Point  friend operator - (Point const& p1,Point const& p2){        return Point(p2.x-p1.x,p2.y-p1.y);    }    double friend operator * (Point const& p1,Point const& p2){        return p1.x*p2.y - p1.y*p2.x;    }    void read() { scanf("%lf%lf",&x,&y); }    void make(double _x,double _y) {x=_x;y=_y;}};struct Segment {    Point p1,p2;};typedef Segment Seg;int InSide(Seg s1,Seg s2,Point p){    if(dcmp(p.y-s1.p1.y)>0 || dcmp(p.y-s1.p2.y)<0 ) return 0;    if(dcmp((s1.p2-p)*(p-s1.p1))>=0 && dcmp((s2.p2-p)*(p-s2.p1))<=0) return 1;    return 0;}int n,m;Seg s[nMax];Point p[nMax];double ui,li;int ans[nMax];int first = 1;int Bserch(Point p){    int l=0,r=n;    int ans = 0;    while(l<=r){        int mid = (l+r)>>1;        if(dcmp((s[mid].p2-p)*(p-s[mid].p1)) >= 0) ans = mid,l = mid+1;        else r = mid-1;    }    return ans;}int main(){#ifndef ONLINE_JUDGEfreopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);#endif    while(~scanf("%d",&n),n){        if(first) first = 0;        else puts("");        scanf("%d",&m);        s[0].p1.read();s[n+1].p2.read();        s[0].p2.make(s[0].p1.x,s[n+1].p2.y);        s[n+1].p1.make(s[n+1].p2.x,s[0].p1.y);        for(int i=1;i<=n;i++) {            scanf("%lf%lf",&ui,&li);            s[i].p1.make(ui,s[0].p1.y);            s[i].p2.make(li,s[0].p2.y);        }        for(int i=1;i<=m;i++) p[i].read();        memset(ans,0,sizeof(ans));        for(int i=1;i<=m;i++) {            ans[Bserch(p[i])]++;        }        for(int i=0;i<=n;i++) printf("%d: %d\n",i,ans[i]);    }    return 0;}

仅为纪念貌似第一道计算几何，虽然好像不是特别难。。。

纯暴力的手段是行不通的，对于每个点我们二分查找它所在的空间。。