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poj 4048 Chinese Repeating Crossbow 线段规范相交的判断

热度:53   发布时间:2024-01-19 05:50:13.0

题意:

给n条线段和初始点,求初始点发出的射线最多能穿过多少条线段(有交点就算穿过)。

分析:

枚举线段端点,判断线段是否规范相交,需要严密的模板。

代码:

//poj 4048
//sep9
#include <iostream>
using namespace std;
typedef long long ll;
const ll maxN=2000;
const ll maxL=40028;struct P
{ll x,y;
}a[maxN],b[maxN],p;ll tx,ty;
int n;ll det(P a,P b,P c)
{ll x1=b.x-a.x;ll y1=b.y-a.y;ll x2=c.x-a.x;ll y2=c.y-a.y;return x1*y2-x2*y1;
}int get_sign(ll x)
{if(x==0) return 0;return x>0?1:-1;
}ll mymax(ll a,ll b)
{return a>b?a:b;
}
ll mymin(ll a,ll b)
{return a>b?b:a;
}
int between(P mid,P a,P b)
{return mid.x<=max(a.x,b.x)&&mid.y<=max(a.y,b.y)&&mid.x>=min(a.x,b.x)&&mid.y>=min(a.y,b.y); 
}
bool crossed(P a,P b,P c,P d)
{int d1,d2,d3,d4;d1=get_sign(det(a,b,c));d2=get_sign(det(a,b,d));d3=get_sign(det(c,d,a));d4=get_sign(det(c,d,b));	 if((d1^d2)==-2&&(d3^d4)==-2)return true;if((d1==0&&between(c,a,b))||(d2==0&&between(d,a,b))||(d3==0&&between(a,c,d))||(d4==0&&between(b,c,d)))return true;return false;
}int process(P dir)
{int cnt=0;dir.x=p.x+(dir.x-p.x)*10000;dir.y=p.y+(dir.y-p.y)*10000;if(dir.x==p.x&&dir.y==p.y) return 0;for(int i=1;i<=n;++i)if(crossed(p,dir,a[i],b[i]))++cnt;return cnt;
}int main()
{int i,cases;scanf("%d",&cases);while(cases--){scanf("%d",&n);for(i=1;i<=n;++i)scanf("%lld%lld%lld%lld",&a[i].x,&a[i].y,&b[i].x,&b[i].y);scanf("%lld%lld",&p.x,&p.y);			int ans=0;for(i=1;i<=n;++i){ans=max(ans,process(a[i]));ans=max(ans,process(b[i]));}printf("%d\n",ans);}return 0;
} 


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