uva 662 - Fast Food
| Fast Food |
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 restaurent 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 (these are the distances measured from the company's headquarter, which happens to be at the same highway). Furthermore, a number 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
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 , , . Following this will n lines containing one integer each, giving the positions d i 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
这道题难度其实不大，仔细分析就能想到dp[i][j]表示前i个餐厅设置j个仓库最小代价。
假设第j个仓库管辖范围[k,i]，这里说明一点，每个仓库的管辖范围肯定是一段连续餐厅区间，否则不可能最优，这个留个你们去证明。
那么dp[i][j]=max{dp[k-1][j-1]+dis[k][i]}，j<=k<=i,这里枚举i,j,k是必须的，复杂度已达O（kn^2)，因此dis数组必须在O(1)的时间内得到。
dis[i][j]说白了是在第i个餐馆到第j个餐馆之间选个位置建个仓库使代价最小，这个显然是中位数。不明白的可以去翻翻中位数的性质：到其他数的距离之和最小。
求dis[i][j]，枚举i,j复杂度已达O（n^2)，因此每个dis[i][j]都必须在O（1）时间里得到,可以再来次dp，dis[i][j]=dis[i][j-1]+a[j]-a[i+j>>1];
或者可以观察一下规律：把所有距离和累加，会发现其实是若干个a[i+j>>1]和两端段区间和相加减。两种方法都可以。这里用第一种了！
代码：
#include<cstdio>
#include<iostream>
#define Maxn 210
using namespace std;
int a[Maxn],dis[Maxn][Maxn],dp[Maxn][40],path[Maxn][40];
const int inf=1<<30;
void print(int n,int m){
if(m==0) return;
int s=path[n][m]/201,e=path[n][m]%201;
print(s-1,m-1);
printf("Depot %d at restaurant %d serves restaurant",m,s+e>>1);
if(s==e) printf(" %d\n",s);
else printf("s %d to %d\n",s,e);
}
int main()
{
int n,m,cas=1;
while(~scanf("%d%d",&n,&m),n){
for(int i=1;i<=n;i++){
scanf("%d",a+i);
dis[i][i]=0;
}
for(int l=1;l<n;l++)
for(int i=1,j=1+l;j<=n;i++,j++)
dis[i][j]=dis[i][j-1]+a[j]-a[i+j>>1];
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
dp[i][j]=inf;
for(int i=1;i<=n;i++){
dp[i][1]=dis[1][i];
path[i][1]=201+i;
}
for(int i=2;i<=n;i++)
for(int j=2;j<=min(i,m);j++)
for(int k=j;k<=i;k++){
if(dp[k-1][j-1]+dis[k][i]<dp[i][j]){
dp[i][j]=dp[k-1][j-1]+dis[k][i];
path[i][j]=k*201+i;
}
}
printf("Chain %d\n",cas++);
print(n,m);
printf("Total distance sum = %d\n\n",dp[n][m]);
}
return 0;
}