广东主任会-中国总经理会新疆分会

刚看过第叁集,好可爱的血小板。即使学医的象征看着某个出戏,但确确实实雅观,就当帮作者复习知识点了。会继续追下去的。话说,里面有壹个只现出一面的长头发的巨噬细胞(白的那些,忘了剧里的现实叫法了)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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宜都首席营业官会、当阳主管会、枝江经理会

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canf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格canf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格 scanf会读空格
   

  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 }

 

奶牛的旅行,1405奶牛旅行 标题描述
Description
农民John的农场里有无数牧区。有的路线连接一些一定的牧区。一片有着连接的牧区称为一…

John想在农场里添加一条路线(注意,恰好一条)。对那条路子有以下限制: 

镇江总监会旗下包涵

1405 奶牛的旅行,1405奶牛旅行

标题叙述 Description

农家John的农场里有诸多牧区。有的路线连接一些一定的牧区。一片有着连接的牧区称为三个牧场。可是就方今而言,你能来看至少有三个牧区通过其余路径都不联网。这样,农民John就有八个牧场了。 

John想在农场里添加一条路子(注意,恰好一条)。对那条路子有以下限制: 

2个牧场的直径就是牧场中最远的八个牧区的相距(本题中所提到的有着距离指的都是最短的离开)。考虑如下的有多少个牧区的牧场,牧区用“*”表示,路径用直线表示。每1个牧区都有投机的坐标: 

       15,15 20,15
         D   E
         *——-*
         |   _/|
         | _/ |
         | _/  |
         |/   |
    *——–*——-*
    A    B   C
    10,10 15,10 20,10
以此牧场的直径差不离是12.07106,
最远的多少个牧区是A和E,它们之间的最短路径是A-B-E。 

那里是另两个牧场: 

             *F 30,15
            / 
           _/ 
          _/  
         /   
         *——* 
         G   H
         25,10 30,10
那多个牧场都在John的农场上。John将会在两个牧场中各选三个牧区,然后用一条途径连起来,使得连通后这几个新的更大的牧场有细微的直径。 

留意,借使两条路线中途相交,大家不以为它们是连着的。只有两条路子在同1个牧区相交,大家才认为它们是联网的。 

输入文件包涵牧区、它们分其他坐标,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
输入文件至少包罗多个不接入的牧区。 

请编程找出一条连接多少个例外牧场的不二法门,使得连上那条途径后,那个更大的新牧场有细微的直径。

输入描述 Input Description

第2行: 2个平头N (1 <= N <= 150), 表示牧区数 

第叁,到N+1行: 每行五个整数X,Y (0 <= X ,Y<= 100000),
表示N个牧区的坐标。注意每一个 牧区的坐标都以不一样等的。 

第N+2行到第2*N+1行: 每行包蕴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

唯有一行,包含三个实数,表示所求直径。数字保留5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 萨姆ple Output

22.071068

数据范围及提醒 Data Size & Hint

1s

难题叙述 Description

建邺首席营业官会、江陵总监会、公安首席执行官会、监利总监会、石首首席执行官会、洪湖老板会

       15,15 20,15
         D   E
         *——-*
         |   _/|
         | _/ |
         | _/  |
         |/   |
    *——–*——-*
    A    B   C
    10,10 15,10 20,10
本条牧场的直径几乎是12.07106,
最远的八个牧区是A和E,它们中间的最短路径是A-B-E。 

南平总监会、通山老板会、崇阳COO会、通城高管会、嘉鱼老板会、赤壁主任会

此间是另三个牧场: 

巴尔的摩COO会旗下包罗:

农民John的农场里有不可胜举牧区。有的路线连接一些一定的牧区。一片拥有连接的牧区称为二个牧场。不过就现阶段而言,你能来看至少有多个牧区通过其余路径都不连贯。那样,农民John就有三个牧场了。 

宁德老总会、南漳总经理会、谷城主任会、保康总经理会、枣阳CEO会、宜城老总会

数量范围及提醒 Data Size & Hint

资阳总监会旗下包含:

第1到N+1行: 每行八个整数X,Y (0 <= X ,Y<= 一千00),
表示N个牧区的坐标。注意每一种 牧区的坐标都以不等同的。 

邵阳老板会、郧县老板会、郧西主管会、竹山总监会、竹溪老板会、房县经理会

输入文件包蕴牧区、它们各自的坐标,还有多个之类的相得益彰邻接矩阵: 

仙桃高管会、天门首席执行官会

  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
输入文件至少蕴涵多少个不衔接的牧区。 

建邺高管会旗下包涵:

瞩目,假诺两条路径中途相交,咱们不觉得它们是过渡的。唯有两条路子在同2个牧区相交,大家才认为它们是联网的。 

来凤老董会、鹤峰主管会

第壹行: 2个平头N (1 <= N <= 150), 表示牧区数 

大庆高管会旗下包蕴:

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