急速pk10_极速pk10网址_极速pk10模式长期稳赚

热门关键词: 急速pk10,极速pk10网址,极速pk10模式长期稳赚

极速pk10网址笔录本人感想,缺憾没假如

来源:http://www.baby3378.com 作者:情感专区 人气:114 发布时间:2019-11-05
摘要:豆蔻梢头经,若无了对错心,可会那么乐观 会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈会哈更会哈更

豆蔻梢头经,若无了对错心,可会那么乐观

会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈更会哈会哈更

本人来测测壹人对于三个摄像可以发布几条批评,哈哈好会还有或然会好会还也许会好会还有只怕会好会还有恐怕会好会还有或者会好会还可能会好会还大概会好会还有只怕会好会还大概会好会还有或许会好会还恐怕会好会还有恐怕会好会还有恐怕会好会还有恐怕会好会还恐怕会哈哈哈哈 好会还有恐怕会好会还大概会好会还有只怕会好会还恐怕会好会还大概会好会还有大概会好会还会好会还只怕会好会还有可能会好会还大概会好会还有也许会好会还或然会好会还有大概会好会还有或者会好会还有大概会哈哈哈哈

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 }

 

红牛的游历,1405水牛游览 标题描述 Description 村民John的农场里有众多牧区。有的路径连接一些特定的牧区。一片有着连接的牧区称为风度翩翩...

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版权文章,未经《短法学》书面授权,严禁转发,违者将被追究法律义务。

极速pk10网址 1

1405 白牛的远足,1405红牛游历

难点陈述 Description

农家约翰的农场里有成千上万牧区。有的路径连接一些特定的牧区。一片有着连接的牧区称为八个牧场。不过就近来来说,你能观察至稀少四个牧区通过此外路线都不连贯。那样,乡里人John就有七个牧场了。 

John想在农场里增加一条路子(注意,正巧一条)。对那条渠道有以下约束: 

一个牧场的直径正是牧场中最远的多个牧区的间距(本题中所提到的全体间隔指的都以最短的偏离)。考虑如下的有5个牧区的牧场,牧区用“*”表示,路线用直线表示。每二个牧区皆有和睦的坐标: 

       15,15 20,15
         D   E
         *-------*
         |   _/|
         | _/ |
         | _/  |
         |/   |
    *--------*-------*
    A    B   C
    10,10 15,10 20,10
本条牧场的直径差非常少是12.07106, 最远的几个牧区是A和E,它们中间的最短路线是A-B-E。 

此地是另一个牧场: 

             *F 30,15
            / 
           _/ 
          _/  
         /   
         *------* 
         G   H
         25,10 30,10
那三个牧场都在John的农场上。John将会在七个牧场中各选多个牧区,然后用一条路线连起来,使得连通后那么些新的更加大的牧场有十分小的直径。 

在意,假若两条门路中途相交,大家不感到它们是连接的。只有两条路子在同二个牧区相交,大家才感觉它们是对接的。 

输入文件包罗牧区、它们各自的坐标,还应该有一个之类的集中民众智慧邻接矩阵: 

  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
输入文件最少蕴涵八个不连贯的牧区。 

请编制程序搜索一条连接八个例外牧场的路线,使得连上这条路径后,这几个更加大的新牧场有细小的直径。

输入描述 Input Description

第1行: 叁个整数N (1 <= N <= 150), 表示牧区数 

第2到N 1行: 每行八个整数X,Y (0 <= X ,Y<= 100000), 表示N个牧区的坐标。注意各样 牧区的坐标都以不均等的。 

第N 2行到第2*N 1行: 每行包涵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

只有生机勃勃行,包罗二个实数,表示所求直径。数字保留五位小数。

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

数据范围及提醒 Data Size & Hint

1s

还是,抑或作者变得服性格很顽强在荆棘塞途或巨大压力面前不屈帖帖你,又会怎么去做

只要,借使再一次能来过何人,又会身伴旁左

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