poj 1556 The Doors

The Doors

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9106		Accepted: 3489

Description

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. 

Output

The output 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.
Sample Input

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

Sample Output

10.00
10.06

分析：还是叉乘，将能走的线路都找出来，能走的路径必定是经过那些线段的边界，再用迪杰斯特拉求最短路。

代码如下：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int n,cnt,inde;
struct Edge{
double x1,x2,y1,y2;
}edge[100]; //cnt
double Map[110][110];
struct Data {
double xi,yi;
}data[100];//inde+1
int vi[100];
double dis[100];
double calc(double x1,double y1,double x2,double y2,double x3,double y3)
{
return (x2-x1)*(y3-y1)-(y2-y1)*(x3-x1);
}
int judge(int a,int b,int x)//判断两点所连线段与其他的都不相交，即题干中两点之间没有障碍物
{
double ans1=calc(data[a].xi,data[a].yi,data[b].xi,data[b].yi,edge[x].x1,edge[x].y1);
double ans2=calc(data[a].xi,data[a].yi,data[b].xi,data[b].yi,edge[x].x2,edge[x].y2);
double ans3=calc(edge[x].x1,edge[x].y1,edge[x].x2,edge[x].y2,data[a].xi,data[a].yi);
double ans4=calc(edge[x].x1,edge[x].y1,edge[x].x2,edge[x].y2,data[b].xi,data[b].yi);
if(ans1*ans2<=0 && ans3*ans4<=0)
return 0;
return 1;
}
int Check(int a,int b)
{
if(data[a].xi==data[b].xi)
return 0;
for(int i=0;i<cnt;i++)
{
if(data[a].xi!=edge[i].x1 && data[a].xi!=edge[i].x2 && data[b].xi!=edge[i].x1 && data[b].xi!=edge[i].x2)
{
if(judge(a,b,i)==0)
return 0;
}
}
return 1;
}

double Distence(int a,int b)
{
return sqrt((data[a].xi-data[b].xi)*(data[a].xi-data[b].xi)+(data[a].yi-data[b].yi)*(data[a].yi-data[b].yi));
}

void Dj()
{
int p;
int n=inde;
for(int i=0;i<=n;i++)
{
dis[i]=Map[0][i];
vi[i]=0;
}
vi[0]=1;
for(int i=0;i<=n;i++)
{
double MIN=0x3f3f3f3f;
for(int j=0;j<=n;j++)
{
if(!vi[j] && dis[j]<MIN)
{
MIN=dis[j];
p=j;
}
}
vi[p]=1;
for(int j=0;j<=n;j++)
{
if(!vi[j] && dis[j]>Map[p][j]+dis[p])
dis[j]=Map[p][j]+dis[p];
}
}
}

int main()
{
double xy[5];
while(scanf("%d",&n))
{
cnt=0; inde=1;
data[0].xi=0; data[0].yi=5;
for(int i=0;i<=100;i++)
for(int j=0;j<=100;j++)
Map[i][j]=0x3f3f3f3f;
if(n==-1)
break;
for(int j=0;j<n;j++)
{
for(int i=0;i<5;i++)
scanf("%lf",&xy[i]);
edge[cnt].x1=xy[0]; edge[cnt].y1=0; edge[cnt].x2=xy[0]; edge[cnt].y2=xy[1]; cnt++;
edge[cnt].x1=xy[0]; edge[cnt].y1=xy[2]; edge[cnt].x2=xy[0]; edge[cnt].y2=xy[3]; cnt++;
edge[cnt].x1=xy[0]; edge[cnt].y1=xy[4]; edge[cnt].x2=xy[0]; edge[cnt].y2=10; cnt++;
for(int i=1;i<=4;i++)
{
data[inde].xi=xy[0]; data[inde].yi=xy[i];
inde++;
}
}
data[inde].xi=10; data[inde].yi=5;
for(int i=0;i<inde;i++)
{
for(int j=i+1;j<=inde;j++)
{
if(Check(i,j))
{
Map[i][j]=Map[j][i]=Distence(i,j);
}
}
}
Dj();
printf("%.2lf\n",dis[inde]);
}
return 0;
}