纯情的血小板,红牛的远足

刚看过第朝气蓬勃集,好可爱的血小板。尽管学医的象征看着有个别出戏,但实在美观,就当帮小编复习知识点了。会三番五次追下去的。话说,里面有三个只现出一面包车型地铁长头发的巨噬细胞(白的十一分,忘了剧里的切实可行叫法了)好帅!!!前面他的戏份会不会多?在打病历,写远远不足140字了。凑字数中……………………………………………………………………………

难题陈诉 Description

山民约翰的农场里有众多牧区。有的路线连接一些特定的牧区。一片具有连接的牧区称为二个牧场。但是就现阶段来讲,你能来看至稀有多个牧区通过任何路线都不连贯。那样,山民John就有八个牧场了。 

John想在农场里增添一条门路(注意,恰巧一条)。对那条门路有以下限定: 

三个牧场的直径正是牧场中最远的五个牧区的偏离(本题中所提到的有着间距指的都是最短的相距)。思量如下的有5个牧区的牧场,牧区用“*”表示,路线用直线表示。每三个牧区都有友好的坐标: 

纯情的血小板,红牛的远足。       15,15 20,15
         D   E
         *——-*
         |   _/|
         | _/ |
         | _/  |
         |/   |
    *——–*——-*
    A    B   C
    10,10 15,10 20,10
以此牧场的直径大概是12.07106,
最远的多个牧区是A和E,它们中间的最短路线是A-B-E。 

那边是另多个牧场: 

             *F 30,15
            / 
           _/ 
          _/  
         /   
         *——* 
         G   H
         25,10 30,10
那五个牧场都在John的农场上。John将会在三个牧场中各选一个牧区,然后用一条门路连起来,使得连通后这几个新的越来越大的牧场有比超小的直径。 

只顾,借使两条渠道中途相交,大家不感觉它们是连着的。独有两条路径在同二个牧区相交,大家才以为它们是联网的。 

输入文件包涵牧区、它们各自的坐标,还应该有贰个之类的舍短取长邻接矩阵: 

  A B C D E F G H 
A 0 1 0 0 0 0 0 0
B 1 0 1 1 1 0 0 0
C 0 1 0 0 1 0 0 0
D 0 1 0 0 1 0 0 0
E 0 1 1 1 0 0 0 0
F 0 0 0 0 0 0 1 0
G 0 0 0 0 0 1 0 1
H 0 0 0 0 0 0 1 0
输入文件最少蕴含八个不联网的牧区。 

请编制程序寻觅一条连接八个差别牧场的门路,使得连上那条门路后,那几个更加大的新牧场有小小的直径。

输入描述 Input Description

第1行: 一个整数N (1 <= N <= 150), 表示牧区数 

第2到N+1行: 每行四个整数X,Y (0 <= X ,Y<= 100000),
表示N个牧区的坐标。注意每个 牧区的坐标都是不相仿的。 

第N+2行到第2*N+1行: 每行满含N个数字(0或1)
表示如上文描述的对称邻接矩阵。

出口描述 Output Description

唯有风度翩翩行,包罗二个实数,表示所求直径。数字保留八个人小数。

样例输入 萨姆ple Input

8
10 10
15 10
20 10
15 15
20 15
30 15
25 10
30 10
01000000
10111000
01001000
01001000
01110000
00000010
00000101
00000010

样例输出 Sample Output

22.071068

数量范围及提醒 Data Size & Hint

1s

分类标签 Tags 点此实行 

 

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  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<queue>
  5 #include<cmath>
  6 using namespace std;
  7 double maxn=1e12;
  8 double x[1001];
  9 double y[1001];
 10 double cd(int i,int j)
 11 {
 12     return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
 13 }
 14 double map[1001][1001];
 15 double m[10001];//点i可以到达的最大距离 
 16 int main()
 17 {
 18     int n;
 19     scanf("%d",&n);
 20     for(int i=1;i<=n;i++)
 21     {
 22         scanf("%lf%lf",&x[i],&y[i]);
 23     }
 24     for(int i=1;i<=n;i++)
 25     {
 26         for(int j=1;j<=n;j++)
 27         {
 28             char kk;
 29             //scanf("%c",&kk);
 30             cin>>kk;
 31             if(kk=='1')
 32             {
 33                 double tmp=cd(i,j);
 34                 map[i][j]=tmp;
 35             }
 36             else
 37             {
 38                 map[i][j]=maxn;
 39             }
 40         }
 41     }
 42     for(int k=1;k<=n;k++)
 43     {
 44         for(int i=1;i<=n;i++)
 45         {
 46             for(int j=1;j<=n;j++)
 47             {
 48                 if(i!=j&&i!=k&&j!=k)
 49                 {
 50                     if(map[i][k]<maxn-1&&map[k][j]<maxn-1)
 51                     {
 52                         if(map[i][j]>map[i][k]+map[k][j])
 53                         {
 54                             map[i][j]=map[i][k]+map[k][j];
 55                         }
 56                     }
 57                     
 58                 }
 59                 
 60             }
 61         }
 62     }
 63     memset(m,0,sizeof(m));
 64     for(int i=1;i<=n;i++)
 65     {
 66         for(int j=1;j<=n;j++)
 67         {
 68             if(map[i][j]<maxn-1)
 69             {
 70                 if(m[i]<map[i][j])
 71                 {
 72                     m[i]=map[i][j];
 73                 }
 74             }
 75             
 76         }
 77     } 
 78     double r2=maxn;
 79     for(int i=1;i<=n;i++)
 80     {
 81         for(int j=1;j<=n;j++)
 82         {
 83             if(map[i][j]>maxn-1&&i!=j)
 84             {
 85                 double tmp=cd(i,j);
 86                 double nn=m[i]+m[j]+tmp;
 87                 if(r2>nn)
 88                 r2=nn;
 89             }
 90         }
 91     }
 92     double r1=-1;
 93     for(int i=1;i<=n;i++)
 94     {
 95         
 96         if(m[i]>r1)
 97         r1=m[i];
 98     }
 99     printf("%.6lf",max(r1,r2));
100     return 0;
101 }

 

 

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