妄迹世所逐,山民John就有八个牧场了

2020-03-12 作者:威尼斯app官网下载   |   浏览(89)

    《晨诣超师范学院读禅经》  笔者:柳河东

1405 白牛的远足,1405白牛游览

难点汇报 Description

同乡约翰的农场里有过多牧区。有的路径连接一些特定的牧区。一片有着连接的牧区称为叁个牧场。但是就现阶段来讲,你能观望至罕有八个牧区通过其余路线都不联网。那样,山民John就有多个牧场了。 

John想在农场里增多一条门路(注意,适逢其会一条卡塔尔。对那条路径有以下约束: 

二个牧场的直径正是牧场中最远的八个牧区的相距(本题中所提到的保有间距指的都以最短的离开State of Qatar。寻思如下的有5个牧区的牧场,牧区用“*”表示,路线用直线表示。每三个牧区都有自个儿的坐标: 

       15,15 20,15
         D   E
         *-------*
         |   _/|
         | _/ |
         | _/  |
         |/   |
    *--------*-------*
    A    B   C
    10,10 15,10 20,10
其一牧场的直径大约是12.07106, 最远的八个牧区是A和E,它们中间的最短路线是A-B-E。 

那边是另一个牧场: 

             *F 30,15
            / 
           _/ 
          _/  
         /   
         *------* 
         G   H
         25,10 30,10
那三个牧场都在John的农场上。约翰将会在七个牧场中各选八个牧区,然后用一条门路连起来,使得连通后这么些新的越来越大的牧场有小小的直径。 

瞩目,借使两条路径中途相交,我们不认为它们是连着的。唯有两条路子在同三个牧区相交,大家才以为它们是对接的。 

输入文件满含牧区、它们分其他坐标,还应该有叁个之类的集思广益邻接矩阵: 

  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
输入文件最少包蕴多少个不接入的牧区。 

请编制程序寻觅一条连接五个不等牧场的不二诀窍,使得连上那条渠道后,那么些更加大的新牧场有细小的直径。

输入描述 Input Description

第1行: 三个卡尺头N (1 <= N <= 150卡塔尔, 表示牧区数 

第2到N+1行: 每行多少个整数X,Y (0 <= X ,Y<= 100000卡塔尔国, 表示N个牧区的坐标。注意每种 牧区的坐标都是不相同的。 

第N+2行到第2*威尼斯app官网下载,N+1行: 每行富含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

独有一行,包蕴叁个实数,表示所求直径。数字保留七个人小数。

样例输入 Sample 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

    汲井漱寒齿,清心拂尘性格很顽强在千难万险或巨大压力面前不屈。

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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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    真源了无取,妄迹世所逐。

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