Fast Food
| Time Limit: 1000MS |
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Memory Limit: 10000K |
| Total Submissions: 1597 |
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Accepted: 553 |
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Special Judge |
Description
The fastfood chain McBurger owns several restaurants along a highway. Recently, they have decided to build several depots along the highway, each one located at a restaurant and supplying several of the restaurants with the needed ingredients. Naturally, these depots should be placed so that the average distance between a restaurant and its assigned depot is minimized. You are to write a program that computes the optimal positions and assignments of the depots.
To make this more precise, the management of McBurger has issued the following specification: You will be given the positions of n restaurants along the highway as n integers d1 < d2 < ... < dn (these are the distances measured from the company's headquarter, which happens to be at the same highway). Furthermore, a number k (k <= n) will be given, the number of depots to be built.
The k depots will be built at the locations of k different restaurants. Each restaurant will be assigned to the closest depot, from which it will then receive its supplies. To minimize shipping costs, the total distance sum, defined as
n
∑ |di - (position of depot serving restaurant i)|
i=1
must be as small as possible.
Write a program that computes the positions of the k depots, such that the total distance sum is minimized.
Input
The input file contains several descriptions of fastfood chains. Each description starts with a line containing the two integers n and k. n and k will satisfy 1 <= n <= 200, 1 <= k <= 30, k <= n. Following this will n lines containing one integer each, giving the positions di of the restaurants, ordered increasingly.
The input file will end with a case starting with n = k = 0. This case should not be processed.
Output
For each chain, first output the number of the chain. Then output an optimal placement of the depots as follows: for each depot output a line containing its position and the range of restaurants it serves. If there is more than one optimal solution, output any of them. After the depot descriptions output a line containing the total distance sum, as defined in the problem text.
Output a blank line after each test case.
Sample Input
6 3
5
6
12
19
20
27
0 0
Sample Output
Chain 1
Depot 1 at restaurant 2 serves restaurants 1 to 3
Depot 2 at restaurant 4 serves restaurants 4 to 5
Depot 3 at restaurant 6 serves restaurant 6
Total distance sum = 8
Source
【題目大意】
一條公路上有n個旅館，選出其中k個設置倉庫，一個倉庫可服務若干個旅館，一個旅館只需一個倉庫服務。問在哪幾個旅館設置倉庫，每個倉庫服務哪些旅館，可使得旅館到倉庫的總距離最小，并求出總距離（長理只要求求最后一步）。
【數(shù)據(jù)范圍】
1 <= n <= 200, 1 <= k <= 30, k <= n
【解題思想】
1、此題屬于明顯動態(tài)規(guī)劃題，關鍵點是找狀態(tài)轉(zhuǎn)移方程。
2、可以用sum[i][j]表示前i個旅館，設置j個倉庫得到的距離和最小值，那么sum[n][k]即為所求。
3、找sum[i][j]的子結(jié)構(gòu)，假設前j-1個倉庫服務第1個到第k個旅館，則最后一個倉庫服務第k+1個到第i個旅館。
4、可以用one[i][j]表示一個倉庫服務第i個到第j個旅館，到這個倉庫距離和的最小值。
5、則得到狀態(tài)轉(zhuǎn)移方程：sum[i][j]=min(sum[k][j-1]+one[k+1][i]) (j-1<=k<=i-1，min表示所有k取值得到的值中的最小值)。
6、問題轉(zhuǎn)換為了求one[i][j]，即在第i到第j家旅館中設置一個倉庫的總距離。
7、假設i到j共有奇數(shù)家旅館，我們嘗試將倉庫放置在中間旅館，即旅館(i+j)/2，假設將倉庫左移距離x，則右半邊所有旅館到倉庫距離均加x，而只有部分左半邊旅館距離減少了x，剩下的減少均小于x，甚至不減少。因此可以得到，將倉庫從中間位置左移到任何位置總距離都會增加，右移同理，因此倉庫放到旅館(i+j)/2最合適。
8、假設i到j共有偶數(shù)家旅館，容易得到將倉庫放到(i+j-1)/2和(i+j+1)/2得到的總距離相等（對稱性），若將倉庫放到(i+j-1)/2，并左移，則用7相似的想法可得知總距離增大，右移情況同理，由此得知倉庫放到(i+j-1)/2這個位置即可滿足總距離最小。
9、由7、8得到one[i][j]實際上時將倉庫放到(i+j)/2取整位置可得到最小的總距離。
10、數(shù)據(jù)范圍較小，我們可以計算出一切one[i][j]的組合。
11、由于poj還要求輸出在哪幾個旅館設置倉庫，每個倉庫服務哪些旅館，因此還需要存儲動態(tài)規(guī)劃路徑。
12、可用at[i][j],from[i][j],to[i][j]分別表示sum[i][j]得到最小值時最后一個倉庫的位置、服務的起始位置和服務的終止位置。
13、通過遞歸輸出結(jié)果。
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int INF=100000000;
int r[300],sum[300][40],one[300][300];
int from[300][40],to[300][40],at[300][40];
int output(int i,int j)
{
if(j<=0||i<=0)return 1;
int num=output(from[i][j]-1,j-1);
printf("Depot %d at restaurant %d serves ",num,at[i][j]);
if(from[i][j]==to[i][j])printf("restaurant %d\n",from[i][j]);
else printf("restaurants %d to %d\n",from[i][j],to[i][j]);
return num+1;
}
int main()
{
int n,K,i,j,k,middle;
int iCase=0;
while(scanf("%d%d",&n,&K)!=EOF)
{
iCase++;
if(n==0&&K==0)break;
for(i=1;i<=n;i++)scanf("%d",&r[i]);
memset(one,0,sizeof(one));
memset(sum,0,sizeof(sum));
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
middle=(i+j)/2;
for(k=i;k<middle;k++)one[i][j]+=r[middle]-r[k];
for(k=middle+1;k<=j;k++)one[i][j]+=r[k]-r[middle];
}
}
for(i=1;i<=n;i++)sum[i][0]=INF;
for(i=1;i<=n;i++)
{
for(j=1;j<=i&&j<=K;j++)
{
sum[i][j]=INF;
for(k=j-1;k<=i-1;k++)
{
int tmp=sum[k][j-1]+one[k+1][i];
if(tmp<sum[i][j])
{
sum[i][j]=tmp;
from[i][j]=k+1;
to[i][j]=i;
at[i][j]=(k+1+i)/2;
}
}
}
}
printf("Chain %d\n",iCase);
output(n,K);
printf("Total distance sum = %d\n\n",sum[n][K]);
}
return 0;
}
文章來源:
http://www.cnblogs.com/kuangbin/archive/2011/11/12/2246407.html