poj2318 水题（二分+叉积）

题目链接：http://poj.org/problem?id=2318

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 5105;
const int maxe = 20000;
const int INF = 0x3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);

struct Point{
    double x,y;
    Point(double x=0, double y=0) : x(x),y(y){ }    //构造函数
};
typedef Point Vector;

Vector operator + (Vector A , Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A , Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (Vector A , double p){return Vector(A.x*p,A.y*p);}
Vector operator / (Vector A , double p){return Vector(A.x/p,A.y/p);}

bool operator < (const Point& a,const Point& b){
    return a.x < b.x ||( a.x == b.x && a.y < b.y);
}
int dcmp(double x){
    if(fabs(x) < eps) return 0;
    else              return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b){
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
double Cross(Vector A, Vector B)  { return A.x*B.y - A.y * B.x; }
double Length(Vector A)    { return sqrt(Dot(A,A)); }


/******************************分割线*******************************/

Point P[maxn][2];
int N,M;

int main()
{
   //freopen("E:\\acm\\input.txt","r",stdin);
   while(scanf("%d",&N)==1 && N){
        scanf("%d",&M);
        double x1,y1,x2,y2;
        scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
        P[0][0] = Point(x1,y2);   P[0][1] = Point(x1,y1);
        P[N+1][0] = Point(x2,y2); P[N+1][1] = Point(x2,y1);
        for(int i=1;i<=N;i++){
            double Ui,Li;
            scanf("%lf %lf",&Ui,&Li);
            P[i][0] = Point(Li,y2);  P[i][1] = Point(Ui,y1);
        }
        int num[maxn];
        memset(num,0,sizeof(num));
        for(int i=1;i<=M;i++){
            Point A;
            scanf("%lf %lf",&A.x,&A.y);
            int L=0, R=N+1;
            while(L < R-1){
                int mid = L + (R-L)/2;
                if(dcmp(Cross(P[mid][1]-P[mid][0],A-P[mid][0])) > 0)  R = mid;
                else                                                  L = mid;
            }
            num[L]++;
        }
        for(int i=0;i<=N;i++){
            printf("%d: %d\n",i,num[i]);
        }
        printf("\n");
   }
}