UVA 216 - Getting in Line

216 - 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

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：把所有电脑连成一条线，使得所用的电缆总长度最小。
两台电脑之间由一条缆线连接，缆线的长度除了这两点间的直线长度，还要额外加上16米长。

一、暴力法

不同的方案就是不同的连接顺序，暴力枚举所有的方案，求出所有方案所用的电缆长度，求出一种长度最短的方案。设每台电脑的编号依次为0,1,2，……，n-1，根据连接顺序，利用STL里面的next_permutation函数求出0~n-1的全排列，每一种排列对应一种连接方案。因为最大只有8台电脑，所以数据量比较小，可以实现。

#include<stdio.h>

#include<math.h>

#include<string.h>

#include<algorithm>

using namespace std;

struct cable

{

    double x,y;

}a[10]; /*保存坐标*/

int b[10],c[10];

double length(double x1,double y1,double x2,double y2) /*求两点之间距离*/

{

    double L=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));

    return L;

}

int main()

{

    double sum,min,l;

    int n,i,cases=0;

    while(~scanf("%d",&n)&&n)

    {

        for(i=0;i<n;i++)

        {

            scanf("%lf%lf",&a[i].x,&a[i].y);

            b[i]=i;

        }

        min=2147483645;

        do

        {

            sum=0;

            for(i=0;i<n-1;i++)

            {

                l=length(a[b[i]].x,a[b[i]].y,a[b[i+1]].x,a[b[i+1]].y)+16;

                sum+=l;

            }

            if(sum<min)

            {

                min=sum;

                for(i=0;i<n;i++)

                   c[i]=b[i];   //可用 memcpy(c,b,sizeof(b)) 代替

            }

        }while(next_permutation(b,b+n));

        printf("**********************************************************\n");

        printf("Network #%d\n",++cases);

        for(i=0;i<n-1;i++)

        {

            l=length(a[c[i]].x,a[c[i]].y,a[c[i+1]].x,a[c[i+1]].y)+16;

            printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n",a[c[i]].x,a[c[i]].y,a[c[i+1]].x,a[c[i+1]].y,l);

        }

        printf("Number of feet of cable required is %.2lf.\n",min);

    }

    return 0;

}

因为next_permutation是C++里面的函数，所以提交时要选择C++语言，不然会编译错误。

二、回溯法

暴力很容易想到，但是不太灵活，也并不是所有都适用的。

而回溯法是更常用的方法，也更加灵活，更难掌握。

回溯法就是深搜（DFS）的变形。一般深搜是要访问所有的解答树的，而回溯也是把问题分成若干步骤并递归求解，但是如果当前步骤已经不是最佳选择的

话，就不继续递归下去，而是返回上一及的递归调用。这样就可以节省很多的时间，而不必徒劳去访问那些“不归路”。

#include<stdio.h>

#include<string.h>

#include<math.h>

struct cable

{

	double x,y;

}a[10];

int b[10],c[10],vis[10],n;

double min,sum,l;

double length(double x1,double y1,double x2,double y2)

{

	double L=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))+16;

	return L;

}

void dfs(int cur,double sum)

{

	int i;

	if(cur==n)

	{

		if(sum<min)

		{

			min=sum;

			memcpy(c,b,sizeof(b));

		}

		return;

	}

	if(sum>=min) return;

	for(i=0;i<n;i++)

	{

		if(vis[i]) continue;

		vis[i]=1;

		b[cur]=i;

		if(cur==0)

			dfs(cur+1,0);

		else

		{

			l=length(a[b[cur]].x,a[b[cur]].y,a[b[cur-1]].x,a[b[cur-1]].y);

			dfs(cur+1,sum+l);

		}

		vis[i]=0;

	}

}

int main()

{

	int cases=0,i;

	while(~scanf("%d",&n)&&n)

	{

		memset(vis,0,sizeof(vis));

		for(i=0;i<n;i++)

		{

			scanf("%lf%lf",&a[i].x,&a[i].y);

			b[i]=i;

		}

		min=99999999;

		dfs(0,0);

		printf("**********************************************************\n");

		printf("Network #%d\n",++cases);

		for(i=1;i<n;i++)

		{

			l=length(a[c[i-1]].x,a[c[i-1]].y,a[c[i]].x,a[c[i]].y);

			printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n",a[c[i-1]].x,a[c[i-1]].y,a[c[i]].x,a[c[i]].y,l);

		}

		printf("Number of feet of cable required is %.2lf.\n",min);

	}

	return 0;

}