[日常摸鱼]UVA393 The Doors 简单计算几何+最短路

Posted 2020-10-15

The  Boy Next   Doors

题意：给定一个固定大小的房间（$x,y$的范围都是$[0,10]$），有$n$个墙壁作为障碍（都与横坐标轴垂直），每个墙壁都有两扇门分别用四个点来描述，起点终点固定在$(0,5)$和$(10,5)$，求起点到终点的最短路长度，$n<=18$

 

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题解：

我们把每堵墙的每一“段”作为一条线段，对任意两点$u,v$，如果两点间的连线不和其他线段相交，那我们从$u$走到$v$的最短距离就是他们的欧几里得距离，对所有点对都这么做一遍，处理出所有能够直接到达的点之间的距离

如果不能直接从$u$到$v$，那么既然是最短路，那走直线一定比走曲线要好，也就是一定是经过他们中间的某些点然后到达的，然后对所有点跑一次最短路就好了

 

这题范围很小Floyd也能过…

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<cstring>
  4 #include<algorithm>
  5 using namespace std;
  6 
  7 const double precision=10e-6;
  8 
  9 const double INF=(~0u>>2);
 10 
 11 const int N=105;
 12 
 13 struct Point 
 14 {
 15     double x,y;
 16 }p[N];
 17 
 18 struct Line
 19 {
 20     Point a,b;
 21 }l[N];
 22 
 23 int n,cnt_point,cnt_line;
 24 
 25 double dist[N][N];
 26 
 27 inline int dblcmp(double d)
 28 {
 29     if(fabs(d)<precision)
 30         return 0;
 31     return (d>0?1:-1);
 32 }
 33 
 34 inline double det(double x1,double y1,double x2,double y2)
 35 {
 36     return x1*y2-x2*y1;
 37 }
 38 
 39 inline double cross(Point a,Point b,Point c)
 40 {
 41     return det(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y);
 42 }
 43 
 44 inline bool SegCrossSimple(Point a,Point b,Point c,Point d)
 45 {
 46     return (dblcmp(cross(c,a,b))^dblcmp(cross(d,a,b)))==-2&&(dblcmp(cross(a,c,d))^dblcmp(cross(b,c,d)))==-2;
 47 }
 48 
 49 inline double sqr2(double x)
 50 {
 51     return x*x;
 52 }
 53 
 54 inline double dis(int i,int j)
 55 {
 56     return sqrt(sqr2(p[i].x-p[j].x)+sqr2(p[i].y-p[j].y));
 57 }
 58 
 59 inline void addPoint(double x,double y)
 60 {
 61     p[++cnt_point]=(Point){x,y};
 62 }
 63 
 64 inline void addLine(double x1,double y1,double x2,double y2)
 65 {
 66     l[++cnt_line]=(Line){(Point){x1,y1},(Point){x2,y2}};
 67 }
 68 
 69 inline void init()
 70 {
 71     memset(p,0,sizeof(p));
 72     memset(l,0,sizeof(l));
 73     cnt_point=cnt_line=0;
 74     addPoint(0,5);addPoint(10,5);
 75 }
 76 
 77 inline void solve()
 78 {
 79     init();
 80     
 81     double x,y1,y2,y3,y4;
 82     for(register int i=1;i<=n;i++)
 83     {
 84         scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
 85         addPoint(x,y1);addPoint(x,y2);
 86         addPoint(x,y3);addPoint(x,y4);
 87         addLine(x,0,x,y1);addLine(x,y2,x,y3);addLine(x,y4,x,10);
 88     }
 89     
 90     for(register int i=1;i<=cnt_point;i++)dist[i][i]=0;
 91     
 92     for(register int i=1;i<=cnt_point;i++)
 93     {
 94         for(register int j=i+1;j<=cnt_point;j++)
 95         {
 96             bool flag=1;
 97             for(register int k=1;k<=cnt_line;k++)
 98             {
 99                 if(SegCrossSimple(p[i],p[j],l[k].a,l[k].b))
100                 {
101                     flag=0;
102                     break;
103                 }
104             }
105             if(flag)
106                 dist[i][j]=dis(i,j);
107             else
108                 dist[i][j]=INF;
109                 
110             dist[j][i]=dist[i][j];
111         }
112     }
113         
114     for(register int k=1;k<=cnt_point;k++)
115         for(register int i=1;i<=cnt_point;i++)
116             for(register int j=1;j<=cnt_point;j++)if(dist[i][k]!=INF&&dist[k][j]!=INF)
117                 dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
118                 
119     printf("%.2lf\n",dist[1][2]);
120 }
121 
122 int main()
123 {
124     while(scanf("%d",&n)==1)
125     {
126         if(n==-1)break;
127         solve();
128     }
129     return 0;
130 }

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