标签：des style blog java color strong
2 2 0 0 1 2 0 1 2 0 0 1 4 1 2
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，要求路径完全包含在圆的覆盖下。
一开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，终于发现了问题，
把圆与圆相交所有交点找出来，排序，去重，然后枚举所有线段，找出线段与圆的所有交点，排序，去重，然后枚举每一段线段是否被某一个圆覆盖，如果不满足，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 4063 线段与圆相交+最短路,布布扣,bubuko.com
标签：des style blog java color strong
原文地址：http://blog.csdn.net/xianxingwuguan1/article/details/37045007
