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

标签：poj   算法   

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;
} 

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

标签：poj   算法   

原文地址：http://blog.csdn.net/sepnine/article/details/44781057