//以下为原class="tags" href="/tags/BLOG.html" title=blog>blog搬迁过来的内容

【题目大意】：给出一个10*10的平面࿰c;对于平面上的每一个竖的截线都可能会有3面墙࿰c;题目会给出墙的端点坐标。然后要求求出(0࿰c;5)到(10࿰c;5)不穿过墙的最短路。

【解题思路】：枚举两个点࿰c;连接成线段࿰c;判断有木有墙与这题线段相交࿰c;如果没有的话࿰c;就可以前进。数据量比较小࿰c;求出所有直达的点后跑一次floyed就好了。

【代码】：

<code class="language-cpp">#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <cctype>
#include <map>
#include <iomanip>
                   
using namespace std;
                   
#define eps 1e-8
#define pi acos(-1.0)
#define inf 1<<30
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long
#define INF 100000

class="tags" href="/tags/STRUCT.html" title=struct>struct Point
{
    double x,y;
    Point() {}
    Point(double a,double b)
    {
        x=a,y=b;
    }
}point[100][100],point1[100];

class="tags" href="/tags/STRUCT.html" title=struct>struct Line
{
    Point a,b;
    Line() {}
    Line(Point x,Point y)
    {
        a=x,b=y;
    }
}line[10000];

double dis[100][100];
int n,cnt1,cnt2,cnt;
double p,q;


inline int sig(double k)
{
    return k < -eps ? -1 : k > eps;
}

inline double det(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}
inline double xmult(Point o, Point a, Point b)
{
    return det(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);
}

inline int intersect1(Point a, Point b, Point c, Point d)
{
    double s1, s2, s3, s4;
    int d1 = sig(s1 = xmult(a, b, c));
    int d2 = sig(s2 = xmult(a, b, d));
    int d3 = sig(s3 = xmult(c, d, a));
    int d4 = sig(s4 = xmult(c, d, b));
    if ((d1^d2) == -2 && (d3^d4) == -2)
    {
        return 1;
    }
    return 0;
}

inline int intersect(Line u, Line v)
{
    return intersect1(u.a, u.b, v.a, v.b);
}

inline double getDist(Point a, Point b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

int main()
{
    while (~scanf("%d",&n))
    {
        if (n==-1) break;
        point1[0]=Point(0,5);
        cnt1=1;
        cnt2=0;
        for (int i=0; i<=n-1; i++)
        {
            scanf("%lf",&p);
            cnt=0;
            point[i][cnt]=Point(p,0);
            cnt++;
            for (int j=1; j<=4; j++)
            {
                scanf("%lf",&q);
                point[i][cnt]=Point(p,q);
                point1[cnt1]=Point(p,q);
                cnt1++;
                cnt++;
            }
            point[i][cnt]=Point(p,10);
            cnt++;
            line[cnt2]=Line(point[i][0],point[i][1]);
            cnt2++;
            line[cnt2]=Line(point[i][2],point[i][3]);
            cnt2++;
            line[cnt2]=Line(point[i][4],point[i][5]);
            cnt2++;
        }
        /* for (int i=0; i<cnt2; i++)
        {
            cout << line[i].a.x << " " << line[i].a.y << " " << line[i].b.x<< " " <<line[i].b.y<<endl;
        }*/
        point1[cnt1]=Point(10,5);
        cnt1++;
        for (int i=0; i<cnt1; i++)
            for (int j=0; j<cnt1; j++)
                dis[i][j]=INF;
        bool flag;
        for (int i=0; i<cnt1; i++)
        {
            for (int j=i+1; j<cnt1; j++)
            {
                Line tmp;
                tmp=Line(point1[i],point1[j]);
                flag=true;
                for (int l=0; l<cnt2; l++)
                {
                    int k;
                    k=intersect(tmp,line[l]);
                    if (k!=0) {flag=false; break;}
                }
                if (flag==true)
                {
                    dis[i][j]=getDist(point1[i],point1[j]);
                }
            }
        }
        for (int k=0;k<cnt1;k++)
            for (int i=0;i<cnt1;i++)
                if (dis[i][k] < INF)
                for (int j=0;j<cnt1;j++)
                if (dis[k][j] < INF)
                    dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
        printf("%.2lf\n",dis[0][cnt1-1]);

    }
    return 0;
}
code>