Uva393 The Doors.

原題鏈接：點擊打開鏈接

題目大意：在一個（10*10）平面內，求出發點（0，5）到終點（10，5） 的最短距離。

計算幾何的基本題目，利用叉積來判斷線段門與門之間是否能夠相連，然後用dijkstra求最短路經

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iomanip>
using namespace std;

const double maxdist=0x7FFFFFFF;
int wall_num;
int edge_num;
int point_num;
double precision=0.00000001;//用來控制判斷精度
double dist[100];
double point_dist[100][100];
struct edge_node{
  double x1,x2,y1,y2;
}edge[80];
struct point_node{
  double x,y;
}point[100];

/*
 *在邊的集合裏添加邊
 *在點的集合裏添加點
 */
void add_edge(double x1,double y1,double x2,double y2){
  edge[edge_num].x1=x1;
  edge[edge_num].x2=x2;
  edge[edge_num].y1=y1;
  edge[edge_num].y2=y2;
  ++edge_num;
}
void add_point(double x,double y){
  point[point_num].x=x;
  point[point_num].y=y;
  point_num++;
}

/*
 *使用dijkstra求兩點之間
 */
void dijkstra(){
  for (int i=1;i<point_num;i++)
    dist[i]=maxdist;
  bool reach[100];
  memset(reach,true,sizeof(reach));
  for (int i=1;i<=point_num;i++){
    int pos;
    double value=maxdist;
    for (int j=0;j<point_num;j++)
      if (dist[j]<value && reach[j]){
	value=dist[j];
	pos=j;
      }
    reach[pos]=false;
    for (int j=0;j<point_num;j++)
      if (reach[j] && 
	  dist[pos]+point_dist[pos][j]<dist[j]){
	dist[j]=dist[pos]+point_dist[pos][j];
      }
  }
}

/*
 *以下一連串子程序用叉積來判斷兩條線段是否相交
 */
double det(double x1,double y1,double x2,double y2){
  return x1*y2-x2*y1;
}
double cross(point_node a,point_node b,point_node c){
  return det(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y);
}
int cmp(double d){
  if (fabs(d)<precision)
    return 0;
  return (d>0)?1:-1;
}
bool segment_cross_simple(point_node a,point_node b,point_node c,point_node d){
  if (((cmp(cross(a,c,d))^cmp(cross(b,c,d)))==-2)&&
      ((cmp(cross(c,a,b))^cmp(cross(d,a,b)))==-2))
    return true;
  else
    return false;
}
/*
 *解決過程
 */
void solve(){
  /*
   *初始化過程
   */
  memset(edge,0,sizeof(edge));
  memset(point,0,sizeof(point));
  memset(dist,0,sizeof(dist));
  memset(point_dist,0,sizeof(point_dist));
  //點集和邊集清零
  edge_num=0;
  point_num=0;
  add_point(0,5);
  add_point(10,5);
  //增加起點和終點
  for (int i=1;i<=wall_num;i++){
    double x,y1,y2,y3,y4;
    cin >> x >> y1 >> y2 >> y3 >> y4;
    
    //輸入每個牆，並添加牆所對應的邊
    add_edge(x,0,x,y1);
    add_edge(x,y2,x,y3);
    add_edge(x,y4,x,10);
    //添加牆所對應的新增頂點
    add_point(x,y1);
    add_point(x,y2);
    add_point(x,y3);
    add_point(x,y4);
  }
  
  for (int i=0;i<point_num;i++)
    for (int j=0;j<point_num;j++)//枚舉任意兩個點
      if (i!=j){//如果他們不是同一個點
	bool link=true;
	for (int k=0;k<edge_num;k++){
	  point_node lv,lv2;
	  lv.x=edge[k].x1;lv.y=edge[k].y1;
	  lv2.x=edge[k].x2;lv2.y=edge[k].y2;
	  if (segment_cross_simple(point[i],point[j],lv,lv2))
	    link=false;
	}
	if (link)
	  point_dist[i][j]=sqrt(pow(point[i].x-point[j].x,2)+
				pow(point[i].y-point[j].y,2));
	else
	  point_dist[i][j]=maxdist;
      }
  //對於任意兩個頂點求他們之間的路徑

  dijkstra();
  //求出源點開始的dijkstra

  cout << setiosflags(ios::fixed) 
       << setprecision(2) 
       << dist[1] << endl; 
}

/*
 *主過程
 */
int main(){
  cin >> wall_num;
  while (wall_num!=-1){
    solve();
    cin >> wall_num;
  }
}