UVA216 Getting in Line

Description

 

 

 Getting in Line 

Computer networking requires that the computers in the network be linked.

This problem considers a ``linear" network in which the computers are chained together so that each is connected to exactly two others except for the two  computers on the ends of the chain which are connected to only one other  computer. A picture is shown below. Here the computers are the black dots  and their locations in the network are identified by planar coordinates  (relative to a coordinate system not shown in the picture).

Distances  between linked computers in the network are shown in feet.

For various reasons it is desirable to minimize the length of cable used.

Your problem is to determine how the computers should be connected into such  a chain to minimize the total amount of cable needed. In the installation  being constructed, the cabling will run beneath the floor, so the amount of  cable used to join 2 adjacent computers on the network will be equal to the  distance between the computers plus 16 additional feet of cable to connect  from the floor to the computers and provide some slack for ease of  installation.

The picture below shows the optimal way of connecting the computers shown above, and the total length of cable required for this configuration is  (4+16)+ (5+16) + (5.83+16) + (11.18+16) = 90.01 feet.

Input

The input file will consist of a series of data sets. Each data set will  begin with a line consisting of a single number indicating the number of  computers in a network. Each network has at least 2 and at most 8 computers.  A value of 0 for the number of computers indicates the end of input.

After  the initial line in a data set specifying the number of computers in a  network, each additional line in the data set will give the coordinates of a  computer in the network. These coordinates will be integers in the range 0  to 150. No two computers are at identical locations and each computer will  be listed once.

Output

The output for each network should include a line which tells the number of  the network (as determined by its position in the input data), and one line  for each length of cable to be cut to connect each adjacent pair of  computers in the network. The final line should be a sentence indicating the  total amount of cable used.

In listing the lengths of cable to be cut, traverse the network from one end to the other. (It makes no difference at which end you start.) Use a format similar to the one shown in the sample  output, with a line of asterisks separating output for different networks  and with distances in feet printed to 2 decimal places.

Sample Input

6
5 19
55 28
38 101
28 62
111 84
43 116
5
11 27
84 99
142 81
88 30
95 38
3
132 73
49 86
72 111
0

Sample Output

**********************************************************
Network #1
Cable requirement to connect (5,19) to (55,28) is 66.80 feet.
Cable requirement to connect (55,28) to (28,62) is 59.42 feet.
Cable requirement to connect (28,62) to (38,101) is 56.26 feet.
Cable requirement to connect (38,101) to (43,116) is 31.81 feet.
Cable requirement to connect (43,116) to (111,84) is 91.15 feet.
Number of feet of cable required is 305.45.
**********************************************************
Network #2
Cable requirement to connect (11,27) to (88,30) is 93.06 feet.
Cable requirement to connect (88,30) to (95,38) is 26.63 feet.
Cable requirement to connect (95,38) to (84,99) is 77.98 feet.
Cable requirement to connect (84,99) to (142,81) is 76.73 feet.
Number of feet of cable required is 274.40.
**********************************************************
Network #3
Cable requirement to connect (132,73) to (72,111) is 87.02 feet.
Cable requirement to connect (72,111) to (49,86) is 49.97 feet.
Number of feet of cable required is 136.99.
题意：求一笔连接所有点的最短路径
分析：题目中以表明点的数量最多不超过8个，完全可以将所有可能循环一遍，用全排列即可 —— next_permutation()
代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
struct point
{
    double x,y;
};
struct Dis
{
    point a,b;
    double dis;
};
double diss(struct point a,struct point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))+16;
}
int main()
{
    point po[10];Dis ansdis[10],zdis[10];
    int n,i,j,t=1,a[10];double min,sum;
    while(cin>>n)
    {
        if(n==0) break;
        for(i=0;i<n;i++) {cin>>po[i].x>>po[i].y;a[i]=i;}
        min=1<<30;
        do
        {
            for(i=0,j=0,sum=0;i<n-1;i++,j++)
            {
                zdis[j].a=po[a[i]];zdis[j].b=po[a[i+1]];zdis[j].dis=diss(po[a[i]],po[a[i+1]]);
                sum=sum+zdis[j].dis;
            }
            if(sum<min)
            {
                min=sum;
                for(i=0;i<n-1;i++)
                {
                    ansdis[i].a=zdis[i].a;ansdis[i].b=zdis[i].b;ansdis[i].dis=zdis[i].dis;
                }
            }
        }
        while(next_permutation(a,a+n));
        cout << "**********************************************************" << endl;
        cout<<"Network #"<<t<<endl;t++;
        for(i=0;i<n-1;i++)
        {
            printf("Cable requirement to connect (%.0lf,%.0lf) to (%.0lf,%.0lf) is %.2lf feet.\n",ansdis[i].a.x,ansdis[i].a.y,ansdis[i].b.x,ansdis[i].b.y,ansdis[i].dis);
        }
        printf("Number of feet of cable required is %.2lf.\n",min);
    }
    return 0;
}