POJ-1556 The Doors

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;
} 
原文地址:https://www.cnblogs.com/nianheng/p/10010095.html