HDU 4063 线段与圆相交+最短路
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.
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.
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)
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;
}
