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 ofnrestaurants along the highway asnintegers(these are the distances measured from the company’s headquarter, which happens to be at the same highway). Furthermore, a numberwill be given, the number of depots to be built.
Thekdepots will be built at the locations ofkdifferent restaurants. Each restaurant will be assigned to the closest depot, from which it will then receive its supplies. To minimize shipping costs, thetotal distance sum, defined as
must be as small as possible.
Write a program that computes the positions of thekdepots, such that the total distance sum is minimized.
Inputwill satisfy. Following this williof the restaurants, ordered increasingly.
The input file will end with a case starting withn=k= 0. This case should not be processed.
OutputFor 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 Input6 356121920270 0Sample OutputChain 1Depot 1 at restaurant 2 serves restaurants 1 to 3Depot 2 at restaurant 4 serves restaurants 4 to 5Depot 3 at restaurant 6 serves restaurant 6Total distance sum = 8
考虑dp状态为dp[i][j]:前i个酒店建k个仓库的最小代价。dp[i][j]=min(dp[k][j-1]+dis[k+1][i])(j-1<=k<i);关于路径输出设path[i][j]为达到dp[i][j]状态的前一状态的末仓库。易知道path[i][1]=0;其他的就是递归输出了。#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int INF = 0x3f3f3f3f;const int maxn=220;int dis[maxn][maxn];int dp[maxn][50];int path[220][50];int d[220];int n,k,cnt;int cal(int l,int r){int sum=0;int mid=(l+r)>>1;for(int i=l;i<=r;i++)sum+=abs(d[mid]-d[i]);return sum;}void print(int i,int x){if(x==0)return ;print(path[i][x],x-1);printf("Depot %d at restaurant %d serves restaurants %d to %d\n",x,(path[i][x]+1+i)/2,path[i][x]+1,i);}int main(){int cas=1;while(~scanf("%d%d",&n,&k)&&(n+k)){CLEAR(dp,INF);CLEAR(path,0);cnt=1;REPF(i,1,n) scanf("%d",&d[i]);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)dis[i][j]=cal(i,j);for(int i=1;i<=n;i++){dp[i][1]=dis[1][i];path[i][1]=0;}for(int i=1;i<=n;i++){for(int j=2;j<=k;j++){for(int p=j-1;p<i;p++)if(dp[p][j-1]+dis[p+1][i]<dp[i][j]){dp[i][j]=dp[p][j-1]+dis[p+1][i];path[i][j]=p;//上个区间}}}printf("Chain %d\n",cas++);print(n,k);printf("Total distance sum = %d\n\n",dp[n][k]);}return 0;}/**/
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