uva216 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 identi ed 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 oor, 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 oor 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 con guration is (4+16)+ (5+16) + (5.83+16) + (11.18+16) = 90.01
feet.
Input
The input le 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 nal 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.
SampleInput
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
SampleOutput
**********************************************************
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.
题目大意：给定一些电脑的笛卡尔平面直角坐标系坐标，求将这些电脑连成一条链的最小总距离（单位：ft），由于连线问题，每两台电脑之间的连接距离要多16 ft。
思路：全枚举暴搜。
可行性剪枝：当早已枚举完了n台电脑时，剪枝。
最优化剪枝：若当前的累计距离总和超过了已求出的最小距离总和，意味着再搜索也无法得出更优解，则剪枝。
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int ans[8],dans[8],n;
bool vis[8];
struct node{int x,y;}a[8];
double dis[8][8],tot;
inline double cal(int i,int j)
{
return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));
}
void dfs(int cur,double sum)
{
if(sum>=tot)return;
if(cur>=n){
tot=sum;
for(int i=0;i<n;i++)
ans[i]=dans[i];
}
for(int i=0;i<n;i++)
if(!vis[i]){
vis[i]=1;
dans[cur]=i;
double tmp=cur?dis[dans[cur-1]][i]:0;
dfs(cur+1,sum+tmp);
vis[i]=0;
}
}
void Init()
{
for(int i=0;i<n;i++)
scanf("%d%d",&a[i].x,&a[i].y);
for(int i=0;i<n;i++)
for(int j=i;j<n;j++)
dis[i][j]=dis[j][i]=cal(i,j)+16;
tot=(double)0x3f3f3f3f;
}
void out(int &T)
{
puts("**********************************************************");
printf("Network #%d\n",++T);
for(int i=1;i<n;i++){
int p=ans[i-1],q=ans[i];
printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",a[p].x,a[p].y,a[q].x,a[q].y,dis[p][q]);
}
printf("Number of feet of cable required is %.2lf.\n",tot);
}
int main()
{
int T=0;
while(scanf("%d",&n)&&n){
Init();
dfs(0,0);
out(T);
}
return 0;
}
//此程序是首次全程在ubuntu（of linux）环境下编写的，以示纪念;
