奶牛的远足,记录自己感想

刚看过第一集,好可爱的血小板。尽管学医的表示看着有些出戏,但实在雅观,就当帮我复习知识点了。会继续追下去的。话说,里面有一个只出现一面的长头发的巨噬细胞(白的百般,忘了剧里的求实叫法了)好帅!!!后边他的戏份会不会多?在打病历,写不够140字了。凑字数中……………………………………………………………………………

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我来测测一个人对于一个电影可以宣布几条评论,哈哈好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会好会还会哈哈哈哈
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题目叙述 Description

澳门新葡萄京娱乐 1

村民约翰(John)的农场里有成千上万牧区。有的路线连接一些一定的牧区。一片拥有连接的牧区称为一个牧场。然而就现阶段而言,你能观望至少有五个牧区通过其余路径都不连贯。这样,农民约翰就有几个牧场了。 

约翰想在农场里添加一条路线(注意,恰好一条)。对这条路子有以下限制: 

一个牧场的直径就是牧场中最远的六个牧区的相距(本题中所提到的拥有距离指的都是最短的距离)。考虑如下的有5个牧区的牧场,牧区用“*”表示,路径用直线表示。每一个牧区都有和好的坐标: 

       15,15 20,15
         D   E
         *——-*
         |   _/|
         | _/ |
         | _/  |
         |/   |
    *——–*——-*
    A    B   C
    10,10 15,10 20,10
这一个牧场的直径大约是12.07106,
最远的六个牧区是A和E,它们中间的最短路径是A-B-E。 

此处是另一个牧场: 

奶牛的远足,记录自己感想。             *F 30,15
澳门新葡萄京娱乐 ,            / 
           _/ 
          _/  
         /   
         *——* 
         G   H
         25,10 30,10
这几个牧场都在约翰的农场上。约翰(John)将会在四个牧场中各选一个牧区,然后用一条路径连起来,使得连通后这多少个新的更大的牧场有细小的直径。 

瞩目,若是两条路线中途相交,我们不觉得它们是对接的。唯有两条路子在同一个牧区相交,我们才认为它们是连着的。 

输入文件包括牧区、它们分此外坐标,还有一个之类的对称邻接矩阵: 

  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
输入文件至少包括六个不接入的牧区。 

请编程找出一条连接五个不等牧场的门路,使得连上这条路子后,这么些更大的新牧场有很小的直径。

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: 每行六个整数X,Y (0 <= X ,Y<= 100000),
表示N个牧区的坐标。注意每个 牧区的坐标都是不均等的。 

第N+2行到第2*N+1行: 每行包括N个数字(0或1)
表示如上文描述的相辅相成邻接矩阵。

输出描述 Output Description

唯有一行,包括一个实数,表示所求直径。数字保留六位小数。

样例输入 山姆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 山姆(Sam)ple Output

22.071068

数据范围及指示 Data Size & Hint

1s

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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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