Poj2318

 Poj2318
    把一个箱子用不相交的n条直线划分成n+1个板块，标号0-n，洒落m个点，统计这些板块内点的个数
解：主要考察叉积的应用，用每个区间右边那条直线去代表区间，也就是说除了题目中的直线还应该加入一条解析式为x=MAX_RIGHT的直线，每加入一个点在线二分查找离这个点最近且在该点右侧的点。

#include<iostream>#include<cstdio>#include<cstdlib>#define eps 1e-10#define MAXM 5000using namespace std;struct point{double x,y;point(){x=y=0;}point(double _x,double _y){x=_x; y=_y;}};struct line{point s,e;}T[MAXM];//s为上层终结，e为下层终结. point V(point a,point b){return point(b.x-a.x,b.y-a.y);}double Cross(point a,point b){return (a.x*b.y-b.x*a.y);}bool left(int mid,point p){double a; point t;t=V(T[mid].s,p);//printf("%d %lf\n",mid,Cross(V(T[mid].s,T[mid].e),V(T[mid].s,p)));//system("pause");return (Cross(V(T[mid].s,T[mid].e),V(T[mid].s,p))<-eps);}int m,n,x1,y1,x2,y2; int area[MAXM];int main(){//freopen("poj.in","r",stdin);bool start=0;double a,b; int l,r,mid;while(1){memset(area,0,sizeof(area));if (start) printf("\n"); start=1;scanf("%d",&n);if (n==0) break;scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);T[n+1].s.x=T[n+1].e.x=x2;T[n+1].s.y=y1; T[n+1].e.y=y2;for (int i=1;i<=n;i++){scanf("%lf%lf",&T[i].s.x,&T[i].e.x);T[i].s.y=y1; T[i].e.y=y2;}while(m--){scanf("%lf%lf",&a,&b);point p=point(a,b);l=1; r=n+1; bool flag;while (l<r){//printf("%d %d\n",l,r);mid=(l+r)>>1;if(left(mid,p)) r=mid;else l=mid+1;}area[l]++; }for(int i=1;i<=n+1;i++) printf("%d: %d\n",i-1,area[i]);}return 0;}