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1405 奶牛的旅行
阅读量:5097 次
发布时间:2019-06-13

本文共 14743 字,大约阅读时间需要 49 分钟。

题目描述 
Description

农民John的农场里有很多牧区。有的路径连接一些特定的牧区。一片所有连通的牧区称为一个牧场。但是就目前而言,你能看到至少有两个牧区通过任何路径都不连通。这样,农民John就有多个牧场了。 

John想在农场里添加一条路径(注意,恰好一条)。对这条路径有以下限制: 
一个牧场的直径就是牧场中最远的两个牧区的距离(本题中所提到的所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。每一个牧区都有自己的坐标: 
       15,15 20,15
         D   E
         *-------*
         |   _/|
         | _/ |
         | _/  |
         |/   |
    *--------*-------*
    A    B   C
    10,10 15,10 20,10
这个牧场的直径大约是12.07106, 最远的两个牧区是A和E,它们之间的最短路径是A-B-E。 
这里是另一个牧场: 
             *F 30,15
            / 
           _/ 
          _/  
         /   
         *------* 
         G   H
         25,10 30,10
这两个牧场都在John的农场上。John将会在两个牧场中各选一个牧区,然后用一条路径连起来,使得连通后这个新的更大的牧场有最小的直径。 
注意,如果两条路径中途相交,我们不认为它们是连通的。只有两条路径在同一个牧区相交,我们才认为它们是连通的。 
输入文件包括牧区、它们各自的坐标,还有一个如下的对称邻接矩阵: 
  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
输入文件至少包括两个不连通的牧区。 
请编程找出一条连接两个不同牧场的路径,使得连上这条路径后,这个更大的新牧场有最小的直径。

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: 每行两个整数X,Y (0 <= X ,Y<= 100000), 表示N个牧区的坐标。注意每个 牧区的坐标都是不一样的。 
第N+2行到第2*N+1行: 每行包括N个数字(0或1) 表示如上文描述的对称邻接矩阵。

输出描述 
Output Description

只有一行,包括一个实数,表示所求直径。数字保留六位小数。

样例输入 
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

数据范围及提示 
Data Size & Hint

1s

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1 #include
2 #include
3 #include
4 #include
5 #include
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]
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&&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 }

 

 

转载于:https://www.cnblogs.com/zwfymqz/p/6690394.html

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