Toy Storage POJ - 2398

Toy Storage

题目链接：https://vjudge.net/problem/POJ-2398#author=Jawen

题目：

 

 思路：跟toys那题差不多，可看上篇博客，但是这题将bc两点作为一条线，然后将所有的线排序，再来依次判断点是否在线的左边

#include<cmath>
#include<stdio.h>
#include<algorithm>
#include<map>
//#include<bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double inf = 1e100;
const double eps = 1e-6;
const int maxn=1e4+10;

struct Point{
    double x, y;
    Point(double x = 0, double y = 0):x(x),y(y){}
};
typedef Point Vector;
Vector operator + (Vector A, Vector B){
    return Vector(A.x+B.x, A.y+B.y);
}
Vector operator - (Point A, Point 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);
}
double Cross(Vector A, Vector B){
    return A.x*B.y-A.y*B.x;
}
bool ToLeftTest(Point a, Point b, Point c){
    return Cross(b - a, c - b) > 0;
}

struct Line{
    Point c, b;
    Line(){}
    bool operator<(const Line &other)const{
        return this->b.x<other.b.x;
    }
};

int main()
{
    //freopen("text","r",stdin);
    int n;
    while(~scanf("%d",&n)&&n){
        int m;
        double x1,y1,x2,y2;
        Point tt[maxn];
        Line tt1[maxn];
        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        // cout<<n<<" "<<m<<' '<<x1<<" "<<y1<<' '<<endl;
        double b, c;
        int pos = 0,pos1=0;
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &c, &b);
            tt[pos].x = c, tt[pos].y = y1;
            tt1[pos1].c=tt[pos];
            pos++;
            tt[pos].x = b, tt[pos].y= y2;
            tt1[pos1].b=tt[pos];
            pos++;
            pos1++;
        }
        tt[pos].x = x2, tt[pos].y = y1;
        tt1[pos1].c=tt[pos];
        pos++;
        tt[pos].x = x2, tt[pos].y = y2;
        tt1[pos1].b=tt[pos];
        sort(tt1,tt1+pos1);
        map<int,int>mm;
        int t1, t2, box[maxn] = {0};
        for (int i = 0; i < m; ++i) {
            scanf("%d%d", &t1, &t2);
            Point a;
            a.x = t1, a.y = t2;
            int num = 0;
            for (int j = 0; j <= n; ++j) {
                if (ToLeftTest(a, tt1[j].b, tt1[j].c)) {
                    //cout<<"i="<<i<<","<<ToLeftTest(a,tt[j+1],tt[j])<<endl;
                    //cout<<"j="<<j<<endl;
                    num = j;
                    break;
                }
            }
            box[num]++;
        }
        sort(box,box+n+1);
        for(int i=0;i<=n;i++)
            mm[box[i]]++;
        for (int i = 0; i <= n;i++) {
            //printf("(%d,%d)(%d,%d)\n",tt[i].x,tt[i].y,tt[i+1].x,tt[i+1].y);
            if(box[i]!=0)
            {
                printf("Box\n");
                break;
            }
        }
       for(map<int,int>::iterator it=mm.begin();it!=mm.end();it++)
       {
           if(it->first!=0)
              printf("%d: %d\n",it->first,it->second);
       }
    }
    return 0;
}