题意:给出一些墙,不能穿墙,求从起点到终点的最短路
就是最短路啊
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+100;
const double inf=1e10;
struct Point{
double x,y;
Point(double xx=0,double yy=0){
x=xx,y=yy;
}
}a[maxn],s[maxn],t[maxn];
struct Vector{
double x,y;
Vector(double xx=0,double yy=0){
x=xx,y=yy;
}
};
int tt,m,v[maxn],nex[maxn],head[maxn],num=1,vis[maxn],n;
double x,sy,ty,w[maxn],bj[maxn];
queue<int>q;
int dcmp(double x){return fabs(x)<1e-9?0:(x>0?1:-1);}
Vector operator - (Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}
double operator * (Vector a,Vector b){return a.x*b.y-a.y*b.x;}
double len(Vector a){return sqrt(a.x*a.x+a.y*a.y);}
void add(int x,int y,double z){
v[++num]=y;
w[num]=z;
nex[num]=head[x];
head[x]=num;
v[++num]=x;
w[num]=z;
nex[num]=head[y];
head[y]=num;
}
int segment(Point a,Point b,Point c,Point d){
Vector x=b-a,y=d-c;
Vector v1=c-a,v2=d-a;
if(dcmp(x*v1)*dcmp(x*v2)>=0) return 0;
v1=a-c,v2=b-c;
if(dcmp(y*v1)*dcmp(y*v2)>=0) return 0;
return 1;
}
void getdist(){
memset(head,0,sizeof(head));
num=1;
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++){
bool ok=1;
for(int k=1;k<=m;k++)
if(segment(a[i],a[j],s[k],t[k])){
ok=0;
break;
}
if(ok) add(i,j,len(a[i]-a[j]));
}
}
double spfa(){
for(int i=1;i<=n;i++)
bj[i]=inf;
bj[1]=0;
q.push(1);
while(!q.empty()){
int now=q.front();
q.pop();
for(int i=head[now];i;i=nex[i])
if(dcmp(bj[v[i]]-bj[now]-w[i])>0){
bj[v[i]]=bj[now]+w[i];
if(!vis[v[i]])
vis[v[i]]=1,q.push(v[i]);
}
vis[now]=0;
}
return bj[n];
}
int main(){
//freopen(".in","r",stdin);
while(1){
scanf("%d",&tt);
if(tt==-1) break;
n=m=0;
a[++n]=Point(0,5);
for(int i=1;i<=tt;i++){
sy=0,scanf("%lf%lf",&x,&ty);
s[++m]=Point(x,sy),t[m]=Point(x,ty);
a[++n]=Point(x,ty);
scanf("%lf%lf",&sy,&ty);
s[++m]=Point(x,sy),t[m]=Point(x,ty);
a[++n]=Point(x,sy),a[++n]=Point(x,ty);
scanf("%lf",&sy),ty=10;
s[++m]=Point(x,sy),t[m]=Point(x,ty);
a[++n]=Point(x,sy);
}
a[++n]=Point(10,5);
getdist();
printf("%.2lf
",spfa());
}
return 0;
}