hdu 1227(动态规划)
Fast Food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2647 Accepted Submission(s): 1124
Problem 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
![](http://acm.hdu.edu.cn/data/images/1227-1.gif)
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.
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
![](http://acm.hdu.edu.cn/data/images/1227-1.gif)
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.
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 a line containing the total distance sum.
Output a blank line after each test case.
Output a blank line after each test case.
Sample Input
6 3
5
6
12
19
20
27
0 0
Sample Output
Chain 1
Total distance sum = 8
Source
题意:在n个餐馆间建k个仓库,求所有餐馆到仓库和最小
分析:dp[i][j]表示前i个餐厅建j个仓库并且第j个仓库建在i点花费的最少代价
假设第 j-1个仓库建设在 k,那么前j个花费的代价为dp[k][j-1]+cost(k,i)cost(k,j)表示k-j的所有餐馆到仓库花费的最少代价
假设第 j-1个仓库建设在 k,那么前j个花费的代价为dp[k][j-1]+cost(k,i)cost(k,j)表示k-j的所有餐馆到仓库花费的最少代价
这题重要的是预处理
#include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<algorithm> #define N 205 using namespace std; int v[N]; int dp[N][35]; ///dp[i][j]表示前i个餐厅建j个仓库并且第j个仓库建在i点花费的最少代价 ///假设第 j-1个仓库建设在 k,那么前j个花费的代价为dp[k][j-1]+cost(k,i) ///cost(k,j)表示k-j的所有餐馆到仓库花费的最少代价 int main() { int n,k; int t = 1; while(scanf("%d%d",&n,&k)!=EOF,n+k){ for(int i=1;i<=n;i++) { scanf("%d",&v[i]); } for(int i=1;i<=n;i++){ ///必要的预处理,因为如果算第1个仓库的时候没有处理,后面的就算不出来了 int cost=0; for(int j=1;j<=i;j++){ cost+=v[i]-v[j]; } dp[i][1]=cost; } for(int j=2;j<=k;j++){ ///枚举仓库数,1已经算过了 for(int i=j;i<=n;i++){ ///枚举餐馆,从j开始,因为仓库数从j开始 dp[i][j]=9999999999; for(int m=j-1;m<i;m++){ int cost = 0; for(int c = m+1;c<i;c++){ cost += min(v[c]-v[m],v[i]-v[c]); } dp[i][j] = min(dp[i][j],dp[m][j-1]+cost); } } } int ans = 9999999999; for(int i=1;i<=n;i++){ ///还只算到dp[i][k] 后面的餐馆到其距离还未加上去 int cost=0; for(int j=i+1;j<=n;j++){ cost+=v[j]-v[i]; } ans = min(ans,dp[i][k]+cost); } printf("Chain %d\nTotal distance sum = %d\n\n",t++,ans); } }