The DoorsZOJ - 1721
You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length.
Input
The input data for the illustrated chamber would appear as follows.
2
4 2 7 8 9
7 3 4.5 6 7
The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input file will contain at least one such set of data. The end of the data comes when the number of walls is -1.
The output file should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.
1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1
10.00
10.06
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<iomanip>
using namespace std;
const double eps=1e-8;
struct Vector{
double x,y;
Vector(){}
Vector(double x,double y):x(x),y(y){}
};
typedef Vector point;
Vector operator + (Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator - (Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator * (Vector a,Vector b){return Vector(a.x*b.x,a.y*b.y);}
Vector operator / (Vector a,Vector b){return Vector(a.x/b.x,a.y/b.y);}
double dot(Vector a,Vector b){
return a.x*b.x+a.y*b.y;
}
double cross(Vector a,Vector b){
return a.x*b.y-a.y*b.x;
}
struct line{
point p1,p2;
line(){}
line(point p1,point p2):p1(p1),p2(p2){}
};
double G[100][100];
vector<line> lv;
vector<point> pv;
bool is_inter(point pa1,point pa2,point pb1,point pb2){
if(max(pa1.x,pa2.x)<=min(pb1.x,pb2.x)
||max(pb1.x,pb2.x)<=min(pa1.x,pa2.x)
||max(pa1.y,pa2.y)<=min(pb1.y,pb2.y)
||max(pb1.y,pb2.y)<=min(pa1.y,pa2.y))return false;
if(cross(pa2-pa1,pb1-pa1)*cross(pa2-pa1,pb2-pa1)<=0
&&cross(pb2-pb1,pa1-pb1)*cross(pb2-pb1,pa2-pb1)<=0)return true;
else return false;
}
double len(Vector a){
return sqrt(dot(a,a));
}
vector<int> u,v;
vector<double> w;
double dis[100];
int main(){
int n;
cin>>n;
while(n!=-1){
for(int i=0;i<n*4+2+10;i++){
dis[i]=10000;
}
u.clear();v.clear();w.clear();
pv.clear();lv.clear();
pv.push_back(point(0,5));
double X,Y1,Y2,Y3,Y4;
for(int i=0;i<n;i++){
cin>>X>>Y1>>Y2>>Y3>>Y4;
point p1(X,Y1),p2(X,Y2),p3(X,Y3),p4(X,Y4);
point p[4]={p1,p2,p3,p4};
pv.push_back(p1);pv.push_back(p2);pv.push_back(p3);pv.push_back(p4);
point pu(X,10),pd(X,0);
lv.push_back(line(pd,p1));lv.push_back(line(p2,p3));lv.push_back(line(p4,pu));
for(int j=0;j<4;j++){
for(int k=0;k<pv.size()-4;k++){
point pa=p[j],pb=pv[k];
int ok=1;
for(int l=0;l<lv.size()-3;l++){
if(is_inter(pa,pb,lv[l].p1,lv[l].p2))ok=0;
}
if(ok){
u.push_back(k);
v.push_back(i*4+1+j);
w.push_back(len(pa-pb));
}
}
}
}
/*for(int i=0;i<pv.size();i++){
cout<<i<<' '<<pv[i].x<<' '<<pv[i].y<<endl;
}*/
for(int k=0;k<pv.size();k++){
point pa(10,5),pb=pv[k];
int ok=1;
for(int l=0;l<lv.size();l++){
if(is_inter(pa,pb,lv[l].p1,lv[l].p2))ok=0;
}
if(ok){
u.push_back(k);
v.push_back(n*4+1);
w.push_back(len(pa-pb));
}
}
dis[0]=0;
//BF算法
/*for(int i=0;i<u.size();i++){
cout<<i<<' '<<u[i]<<' '<<v[i]<<' '<<w[i]<<endl;
}*/
for(int i=0;i<n*4+2;i++){
int is_renew=0;
for(int j=0;j<u.size();j++){
if(dis[v[j]]>dis[u[j]]+w[j]){
dis[v[j]]=dis[u[j]]+w[j];
is_renew=1;
}
}
if(is_renew==0)break;
}
cout<<fixed<<setprecision(2)<<dis[n*4+1]<<endl;
cin>>n;
}
return 0;
}