poj 2318 TOYS

poj 2318 TOYS

• 思路:这道题很水，只是要知道会使用叉乘来表示点在线的上面还是下面；
  当a.Xmult(b,c) < 0时，表示在线的上面。之后就是二分的时候，不能直接使用mid来ans[mid]++;因为只是确定点在这条线的两边，到底是哪一边，具体还要用tmp来判断;(模板题)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define MS0(a) memset(a,0,sizeof(a))
const int MAXN = 5050;
struct point{
    int x,y;
    point(){}
    point(int _x,int _y){
        x = _x; y = _y;
    }
    long long operator *(const point &b)const{// 叉乘
        return (1LL*x*b.y - 1LL*y*b.x);
    }
    point operator -(const point &b)const{
        return point(x - b.x,y - b.y);
    }
    long long dot(const point &b){    //点乘
        return 1LL*x*b.x + 1LL*y*b.y;
    }
    double dist(const point &b){
        return sqrt(1LL*(x-b.x)*(x-b.x)+1LL*(y-b.y)*(y-b.y));
    }
    long long dist2(const point &b){
        return 1LL*(x-b.x)*(x-b.x)+1LL*(y-b.y)*(y-b.y);
    }
    double len(){
        return sqrt(1LL*x*x+1LL*y*y);
    }
    double point_to_segment(point b,point c)//点a到“线段” bc的距离a.point_to_segment(b,c);
    {
        point v[4];
        v[1] = {c.x - b.x,c.y - b.y};
        v[2] = {x - b.x,y - b.y};
        v[3] = {x - c.x,y - c.y};
        if(v[1].dot(v[2]) < 0) return v[2].len();
        if(v[1].dot(v[3]) > 0) return v[3].len();
        return fabs(1.*(v[1]*v[2])/v[1].len());
    }
    long long Xmult(point b,point c){   // 当a->b与a->c顺时针转时，返回正；
        return (b-*this)*(c-*this);
    }
    void input(){
        scanf("%d%d",&x,&y);
    }
}p[MAXN];

struct Line{
    point s,t;
    Line(){}
    Line(point _s,point _t){
        s = _s,t =_t;
    }
}line[MAXN];
int ans[MAXN];
int main()
{
    int n,m,i,j,x1,y1,x2,y2,kase = 0,U,L;
    while(scanf("%d",&n),n){
        MS0(ans);
        if(kase) puts("");
        else kase++;
        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
        for(i = 1;i <= n;i++){
            scanf("%d%d",&U,&L);
            line[i] = Line(point(U,y1),point(L,y2));
        }
        line[0] = Line(point(x1,y1),point(x1,y2));
        int x,y;
        for(i = 0;i < m;i++){
            scanf("%d%d",&x,&y);
            int l = 0, r = n,mid,tmp;
            while(l <= r){
                mid = l + r >> 1;
                if( point(x,y).Xmult(line[mid].s,line[mid].t) < 0) r = mid-1; //在线的上边
                else tmp = mid,l = mid+1;   //线下的点所在的区域才是改line的标号；
            }
            ans[tmp]++;

        }
        for(i = 0;i <= n;i++){
            printf("%d: %d\n",i,ans[i]);
        }
    }
    return 0;
}