高效算法设计专项：UVa 10535

暴力枚举从起始点引一条到墙的一个端点的射线所穿过的墙的总数即可。复杂度为O(n^2)。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
#define M 510
const double eps=1e-8;
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y){}
};
point operator-(point a,point b){return point(a.x-b.x,a.y-b.y);}
point operator+(point a,point b){return point(a.x+b.x,a.y+b.y);}
point operator*(point a,double p){return point(a.x*p,a.y*p);}
double cross(point a,point b){return a.x*b.y-a.y*b.x;}
double dot(point a,point b){return a.x*b.x+a.y*b.y;}
bool seginter(point a1,point a2,point b1,point b2)
{
    double c1=cross(a2-a1,b1-a1),c2=cross(a2-a1,b2-a1);
    double c3=cross(b2-b1,a1-b1),c4=cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
bool onseg(point p,point a1,point a2)
{
    return dcmp(cross(a1-p,a2-p))==0&&dcmp(dot(a1-p,a2-p))<0;
}
int n;
point s[M],e[M],q;
int main()
{
    freopen("in.txt","r",stdin);
    while(cin>>n&&n)
    {
        for(int i=0;i<n;i++) cin>>s[i].x>>s[i].y>>e[i].x>>e[i].y;
        cin>>q.x>>q.y;
        int ans=0;
        for(int i=0;i<n;i++)
        {
            point tmp;int cnt;
            tmp=q+(s[i]-q)*1e9;
            cnt=0;
            for(int j=0;j<n;j++)
            {
                if(seginter(q,tmp,s[j],e[j])||onseg(s[j],q,tmp)||onseg(e[j],q,tmp))
                    cnt++;
            }
            ans=max(ans,cnt);
            tmp=q+(e[i]-q)*1e9;
            cnt=0;
            for(int j=0;j<n;j++)
            {
                if(seginter(q,tmp,s[j],e[j])||onseg(s[j],q,tmp)||onseg(e[j],q,tmp))
                    cnt++;
            }
            ans=max(ans,cnt);
        }
        cout<<ans<<endl;
    }
    return 0;
}