POJ 1066 Treasure Hunt 线段相交判断

Posted 2020-09-29 Pacify

判断以宝藏的坐标和中点的坐标为线段的点是否与墙相交，求最少相交的墙的数量

中点算出来，枚举中点和墙

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define eps 1e-8
#define INF 1e9
using namespace std;

const int maxn=100;

typedef struct Point
{
    double x,y;
    Point() {};
    Point(double xx,double yy)
    {
        x=xx;
        y=yy;
    }
} Vector;

struct Line
{
    Point p,q;
    Line() {};
    Line(Point pp,Point qq)
    {
        p=pp;
        q=qq;
    }
};

int sgn(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0? -1:1;
}


double crs_prdct(Vector a,Vector b)
{
    return a.x*b.y-b.x*a.y;
}

double dot_prdct(Vector a,Vector b)
{
    return a.x*a.y+b.x*b.y;
}

Point mid_point(Point a,Point b)
{
    return Point((a.x+b.x)/2,(a.y+b.y)/2);
}

Vector operator - (Point a,Point b)
{
    return Vector(a.x-b.x,a.y-b.y);
}

bool operator == (Point a,Point b)
{
    return a.x==b.x && a.y==b.y;
}

bool inter(Line l1,Line l2)
{
    return
        max(l1.p.x,l1.q.x) >= min(l2.p.x,l2.q.x) &&
        max(l2.p.x,l2.q.x) >= min(l1.p.x,l1.q.x) &&
        max(l1.p.y,l1.q.y) >= min(l2.p.y,l2.q.y) &&
        max(l2.p.y,l2.q.y) >= min(l1.p.y,l1.q.y) &&
        sgn(crs_prdct(l2.p-l1.p,l1.q-l1.p))*sgn(crs_prdct(l2.q-l1.p,l1.q-l1.p))<=0 &&
        sgn(crs_prdct(l1.p-l2.p,l2.q-l1.p))*sgn(crs_prdct(l1.q-l2.p,l2.q-l2.p))<=0;
}

bool cmp(Point a,Point b)
{
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}

Line line[maxn];
Point mid[4*maxn];
Point a[4][maxn];

int main()
{
//    freopen("in.txt","r",stdin);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        double x1,y1,x2,y2;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[i]=Line(Point(x1,y1),Point(x2,y2));
        }
        Point p;
        scanf("%lf%lf",&p.x,&p.y);
        a[0][0]=Point(0,0);
        a[1][0]=Point(0,0);
        a[2][0]=Point(0,100);
        a[3][0]=Point(100,0);
        int cnt[4]={1,1,1,1};
        for(int i=0;i<n;i++)
        {
            if(line[i].p.y==0) a[0][cnt[0]++]=Point(line[i].p.x,0);
            if(line[i].q.y==0) a[0][cnt[0]++]=Point(line[i].q.x,0);
            if(line[i].p.x==0) a[1][cnt[1]++]=Point(0,line[i].p.y);
            if(line[i].q.x==0) a[1][cnt[1]++]=Point(0,line[i].q.y);
            if(line[i].p.y==100) a[2][cnt[2]++]=Point(line[i].p.x,100);
            if(line[i].q.y==100) a[2][cnt[2]++]=Point(line[i].q.x,100);
            if(line[i].p.x==100) a[3][cnt[3]++]=Point(100,line[i].p.y);
            if(line[i].q.x==100) a[3][cnt[3]++]=Point(100,line[i].q.y);
        }
        a[0][cnt[0]++]=Point(100,0);
        a[1][cnt[1]++]=Point(0,100);
        a[2][cnt[2]++]=Point(100,100);
        a[3][cnt[3]++]=Point(100,100);
        for(int i=0;i<4;i++)
            sort(a[i],a[i]+cnt[i],cmp);
        int cnt2=0;
        for(int i=0;i<4;i++)
            for(int j=0;j<cnt[i]-1;j++)
                if(!(a[i][j]==a[i][j+1]))
                    mid[cnt2++]=mid_point(a[i][j],a[i][j+1]);
        int ans=INF;
        for(int i=0;i<cnt2;i++)
        {
            int tmp=0;
            for(int j=0;j<n;j++)
                if(inter(Line(p,mid[i]),line[j])) tmp++;
            ans=min(ans,tmp);
        }
        printf("Number of doors = %d \n",ans+1);
    }
    return 0;
}