HDU 4063 线段与圆相交+最短路

本文探讨了在一款飞行游戏中寻找从起点到终点的最短路径问题,特别关注于路径必须完全位于由无线信号源形成的圆覆盖区域内的情况。文中详细介绍了如何通过计算圆与圆之间的交点,并结合线段与圆的交点来确定有效路径的方法。

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Aircraft

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 980    Accepted Submission(s): 228


Problem Description
You are playing a flying game.
In the game, player controls an aircraft in a 2D-space.
The mission is to drive the craft from starting point to terminal point.
The craft needs wireless signal to move.
A number of devices are placed in the 2D-space, spreading signal.
For a device Di, it has a signal radius -- Ri.
When the distance between the craft and Di is shorter or equal to Ri, it(the craft) gets Di's wireless signal.
Now you need to tell me the shortest path from starting point to terminal point.
 

Input
The first line of the input file is a single integer T.
The rest of the test file contains T blocks.
Each block starts with an integer n, followed by n devices given as (xi, yi, Ri).
(xi, yi) is position of Di, and Ri is the radius of its signal range.
The first point is the starting point.
The last point is the terminal point.
T <= 25;
2 <= n <= 20 for most cases;
20 < n <= 25 for several cases, completely random generated.
-1000 <= xi, yi <= 1000 , 1 <= ri <= 1000.
All are integers.
 

Output
For each case, Output "No such path." if the craft can't get to the terminal point.
Otherwise, output a float number, correct the result to 4 decimal places.(as shown in the sample output)
 

Sample Input
2 2 0 0 1 2 0 1 2 0 0 1 4 1 2
 

Sample Output
Case 1: 2.0000 Case 2: No such path.
 


给定n个圆,求从第一个圆到最后一个圆的最短路,要求路径全然包括在圆的覆盖下。

一開始以为圆与圆之间的处理必须经过圆心。无数的wa,苦苦找不到问题。一神牛提供了一组强大的数据,最终发现了问题。

把圆与圆相交全部交点找出来,排序,去重。然后枚举全部线段,找出线段与圆的全部交点,排序。去重,然后枚举每一段线段是否被某一个圆覆盖。假设不满足,直接

跳出,然后就是最短路的处理,连最短路也不会写了,直接堆了一个spfa,300+行代码,总算1Y了,泪目呀。。。。

代码:

/* ***********************************************
Author :rabbit
Created Time :2014/7/3 22:46:38
File Name :2.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-6
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
	if(fabs(x)<eps)return 0;
	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);
}
Point operator / (Point a,double p){
	return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
	return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
	return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a, Point b){
	return a.x*b.x+a.y*b.y;
}
double Length(Point a){
	return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
	return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
	return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
	return a.x*b.y-a.y*b.x;
}
Point vecnit(Point x){
	return x/Length(x);
}
Point normal(Point x){
	return Point(-x.y,x.x)/Length(x);
}
Point Rotate(Point a,double rad){
	return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
	Point u=p-q;
	double t=Cross(w,u)/Cross(v,w);
	return p+v*t;
}
struct Line{
	Point p,v;
	double ang;
	Line(){};
	Line(Point _p,Point _v):p(_p),v(_v){
		ang=atan2(v.y,v.x);
	}
	Point point(double a){
		return p+(v*a);
	}
	bool operator < (const Line &L)const{
		return ang<L.ang;
	}
};
Point GetLineIntersection(Line a,Line b){
	return GetLineIntersection(a.p,a.v,b.p,b.v);
}
bool OnLeft(const Line &L,const Point &p){
	return Cross(L.v,p-L.p)>=0;
}
vector<Point> HPI(vector<Line> L){
	int n=L.size();
	sort(L.begin(),L.end());
	int first,last;
	vector<Point> p(n);
	vector<Line> q(n);
	vector<Point> ans;
	q[last=first=0]=L[0];
	for(int i=1;i<n;i++){
		while(first<last&&!OnLeft(L[i],p[last-1]))last--;
		while(first<last&&!OnLeft(L[i],p[first]))first++;
		q[++last]=L[i];
		if(fabs(Cross(q[last].v,q[last-1].v))<eps){
			last--;
			if(OnLeft(q[last],L[i].p))q[last]=L[i];
		}
		if(first<last)p[last-1]=GetLineIntersection(q[last-1],q[last]);
	}
	while(first<last&&!OnLeft(q[first],p[last-1]))last--;
	if(last-first<=1)return ans;
	p[last]=GetLineIntersection(q[last],q[first]);
	for(int i=first;i<=last;i++)ans.push_back(p[i]);
	return ans;
}
double getarea(vector<Point> p){
	double ans=0;
	int n=p.size();
	for(int i=0;i<n;i++)
		ans+=Cross(p[i],p[(i+1)%n]);
	return fabs(ans)/2;
}
bool getdir(vector<Point> p){
	double ans=0;
	int n=p.size();
	for(int i=0;i<n;i++)
		ans+=Cross(p[i],p[(i+1)%n]);
	if(dcmp(ans)>0)return 1;
	return 0;
}
struct Circle{
	Point c;
	double r;
	Circle(){}
	Circle(Point _c,double _r):c(_c),r(_r){}
	Point point(double a){
		return Point(c.x+cos(a)*r,c.y+sin(a)*r);
	}
};
bool OnSegment(Point p,Point a1,Point a2){
	return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<=0;
}
int getSegCircleIntersection(Line L, Circle C, Point* sol)
{
    Point nor = normal(L.v);
    Line pl = Line(C.c, nor);
    Point ip = GetLineIntersection(pl, L);
    double dis = Length(ip - C.c);
    if (dcmp(dis - C.r) > 0) return 0;
    Point dxy = vecnit(L.v) * sqrt(C.r*C.r-dis*dis);
    int ret = 0;
    sol[ret] = ip + dxy;
    if (OnSegment(sol[ret], L.p, L.point(1))) ret++;
    sol[ret] = ip - dxy;
    if (OnSegment(sol[ret], L.p, L.point(1))) ret++;
    return ret;
}
int getCircleCircleIntersection(Circle c1,Circle c2,vector<Point> &sol){
	double d=Length(c1.c-c2.c);
	if(dcmp(d)==0){
		if(dcmp(c1.r-c2.r)==0)return -1;
		return 0;
	}
	if(dcmp(c1.r+c2.r-d)<0)return 0;
	if(dcmp(fabs(c1.r-c2.r)-d)>0)return 0;
	double a=angle(c2.c-c1.c);
	double da=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
	Point p1=c1.point(a-da),p2=c1.point(a+da);
	sol.push_back(p1);
	if(p1==p2)return 1;
	sol.push_back(p2);
	return 2;
}
bool InCircle(Point x,Circle c){
	return dcmp(c.r-Length(c.c-x))>=0;
}
Point pp[50],p1[10010];
double R[50],dist[1010][1010],dis[1010];
int tot;
int head[1010],tol,vis[1910];
struct Edge{
	int next,to;
	double val;
}edge[1001000];
void addedge(int u,int v,double val){
	edge[tol].to=v;
	edge[tol].next=head[u];
	edge[tol].val=val;
	head[u]=tol++;
}
double spfa(int s,int t){
	memset(vis,0,sizeof(vis));queue<int> q;
	for(int i=0;i<tot;i++){
		if(i==s){
			vis[i]=1;
			dis[i]=0;
			q.push(i);
		}
		else dis[i]=INF;
	}
	while(!q.empty()){
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=head[u];i!=-1;i=edge[i].next){
			int v=edge[i].to;
			if(dis[v]>dis[u]+edge[i].val){
				dis[v]=dis[u]+edge[i].val;
				if(vis[v]==0){
					vis[v]=1;
					q.push(v);
				}
			}
		}
	}
//	cout<<"gg: "<<endl;
	//cout<<"tot="<<tot<<endl;
//	for(int i=0;i<tot;i++)cout<<dis[i]<<" ";cout<<endl;
	if(dis[t]<INF)return dis[t];
	return -1;
}
int main()
{
	int T,n;
//	freopen("data.in","r",stdin);
   //  freopen("data.out","w",stdout);
	scanf("%d",&T);
	for(int t=1;t<=T;t++){
		scanf("%d",&n);
		for(int i=0;i<n;i++)
			scanf("%lf%lf%lf",&pp[i].x,&pp[i].y,&R[i]);
		printf("Case %d: ",t);
		 tot=0;
		for(int i=0;i<n;i++)p1[tot++]=pp[i];
		for(int i=0;i<n;i++)
			for(int j=i+1;j<n;j++){
				vector<Point> sol;
				int ret=getCircleCircleIntersection(Circle(pp[i],R[i]),Circle(pp[j],R[j]),sol);
				for(int k=0;k<ret;k++)
					p1[tot++]=sol[k];
			}
		sort(p1,p1+tot);
		tot=unique(p1,p1+tot)-p1;
		for(int i=0;i<1000;i++){
			dist[i][i]=0;
			for(int j=i+1;j<1000;j++)
				dist[i][j]=dist[j][i]=INF;
		}
		for(int i=0;i<tot;i++)
			for(int j=i+1;j<tot;j++){
				Point tt[100],sol[3];
				int hh=0;
				for(int k=0;k<n;k++){
					int ret=getSegCircleIntersection(Line(p1[i],p1[j]-p1[i]),Circle(pp[k],R[k]),sol);
					for(int d=0;d<ret;d++)
						tt[hh++]=sol[d];
				}
				tt[hh++]=p1[i];
				tt[hh++]=p1[j];
				sort(tt,tt+hh);
				hh=unique(tt,tt+hh)-tt;
				int ff=1;
				for(int d=0;d<hh-1;d++){
					int flag=0;
					for(int e=0;e<n;e++)
						if(InCircle(tt[d],Circle(pp[e],R[e]))&&InCircle(tt[d+1],Circle(pp[e],R[e]))){
							flag=1;break;
						}
					if(!flag){
						ff=0;break;
					}
				}
				if(ff)dist[i][j]=dist[j][i]=Length(p1[i]-p1[j]);
		}
	//	cout<<"tot="<<tot<<endl;
	//	for(int i=0;i<tot;i++)cout<<p1[i].x<<" "<<p1[i].y<<endl;
	//	for(int i=0;i<tot;i++){
	//		for(int j=0;j<tot;j++)cout<<dist[i][j]<<" ";
	//		cout<<endl;
	//	}
		int start,end;
		for(int i=0;i<tot;i++){
			if(p1[i]==pp[0])start=i;
			if(p1[i]==pp[n-1])end=i;
		}
	//	cout<<"han "<<start<<" "<<end<<endl;
		memset(head,-1,sizeof(head));tol=0;
		for(int i=0;i<tot;i++)
			for(int j=i+1;j<tot;j++)
				if(dist[i][j]<INF){
					addedge(i,j,dist[i][j]);
					addedge(j,i,dist[i][j]);
				}
	//	cout<<"tol="<<tol<<endl;
		double ans=spfa(start,end);
		if(ans==-1)puts("No such path.");
		else printf("%.4f\n",ans);
	}
	return 0;
}


HDU(Hangzhou Dianzi University)OJ 中经常涉及到几何计算的问题,其中“判断两条线段是否相交”是一个经典的算法问题。以下是关于如何判断两线段是否相交的基本思路及其实现步骤: ### 判断两条线段相交的核心思想 可以利用向量叉积以及端点位置的关系来确定两条线段是否相交。 #### 具体步骤: 1. **定义基本概念** - 假设两条线段分别为 `AB` 和 `CD`。 - 使用二维平面中的坐标表示各顶点:A(x₁,y₁), B(x₂,y₂),C(x₃,y₃) ,D(x₄,y₄)。 2. **叉积的作用** 叉积可以帮助我们了解两点相对于一条直线的位置关系。 对于三个点 P、Q、R ,我们可以用叉乘 `(Q-P)x(R-P)` 来检测 R 是否在 QP 直线的一侧还是另一侧。 如果结果为正数,则表明顺时针;如果负则逆时针;若等于0则共线。 3. **快速排斥实验** 首先做一个矩形包围盒测试——即检查两个线段所在的小外接矩形是否有重叠区域。如果没有重叠直接判定为不相交。 4. **跨立试验 (Cross-over Test)** 确认每个线段的两端分别位于另一个线段两侧即可认为它们交叉了。这通过上述提到过的叉积运算完成。 5. **特殊情况处理** 包含但不限于如下的几种情况需要单独讨论: - 完全重合的部分; - 存在一个公共端点但并不完全穿过等边缘状况。 6. **代码框架示例(Pseudo code):** ```python def cross_product(p1,p2,p3): return (p2[0]-p1[0])*(p3[1]-p1[1])-(p2[1]-p1[1])*(p3[0]-p1[0]) def on_segment(p,q,r): if ((q[0] <= max(p[0], r[0])) and (q[0] >= min(p[0], r[0])) and (q[1] <= max(p[1], r[1])) and (q[1] >= min(p[1], r[1]))): return True; return False; def do_segments_intersect(A,B,C,D): # 计算四个方向的叉积值 o1 = cross_product(A, C, B) o2 = cross_product(A, D, B) o3 = cross_product(C, A, D) o4 = cross_product(C, B, D) # 标准情况判断 if(o1 !=o2 && o3!=o4): return True # 特殊情况逐一验证... ``` 7. 终结合所有条件得出结论。
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