poj 2318 题解

题意:给出一个箱子四个点的坐标和n个已经排好序的不相交的隔板，隔板不相交，将箱子分隔，再给出m个点(点不在隔板上)，求每个区域内有几个点

多组数据，n<=5000,m<=5000

题解:二分+叉积判断点是否在直线右边

注意:由题中Hint,在箱子右边界上的点算作最后一个区域的点；题中没给出坐标数据范围，直接用了long long

 

 

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define rep(i,a,b) for(int i=a;i<=b;++i)

const double eps=1e-8;

struct point{
    double x,y;
    point(){}
    point (double a,double b): x(a),y(b) {}

    friend point operator + (const point &a,const point &b){
        return point(a.x+b.x,a.y+b.y);
    }

    friend point operator - (const point &a,const point &b){
        return point(a.x-b.x,a.y-b.y);
    }

    friend bool operator == (const point &a,const point &b){
        return (abs(a.x-b.x)<eps && abs(a.y-b.y)<eps);
    }

};

inline double det(point a,point b) {return a.x*b.y-a.y*b.x;}
inline bool line_cross_segment(point s,point t,point a,point b)
{
    return det(s-a,t-a)*det(s-b,t-b)<eps;
}

point s[200][2];
double sx,sy,tx,ty;
int t,n;


bool test(int x,int y)
{
    point ts,tt;
    bool flag=false;
    rep(i,0,1)
        rep(j,0,1) if (!(s[x][i]==s[y][j]))
            {
                flag=true;
                ts=s[x][i];tt=s[y][j];
                rep(k,1,n)
                    if (!line_cross_segment(ts,tt,s[k][0],s[k][1]))
                    {
                        flag=false;
                        break;
                    }
                if (flag) break;
            }
    if (flag) return true;
        else return false;
}

int main()
{
    scanf("%d",&t);
    bool flag;
    while (t--)
    {
        scanf("%d",&n);
        flag=false;
        rep(i,1,n)
            {
                scanf("%lf%lf%lf%lf",&sx,&sy,&tx,&ty);
                s[i][0]=point(sx,sy);s[i][1]=point(tx,ty);
            }
        if (n<=2) {printf("Yes!\n");continue;}
        rep(i,1,n)
            rep(j,i+1,n)
                if(test(i,j)) {flag=true;break;}
        if (flag) printf("Yes!\n");else printf("No!\n");
    }
    return 0;
}