hdu 1227(經典dp)
來源：程序員人生 發布時間：2014-09-29 15:03:00 閱讀次數：3517次
題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1227
Fast Food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2165 Accepted Submission(s): 926
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
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 a line containing the total distance sum.
Output a blank line after each test case.
Sample Input
Sample Output
Chain 1
Total distance sum = 8
Source
Southwestern Europe 1998
題意：一條直線上有n家餐館，現在要在這n家餐館中選出k個來建立倉庫，每家餐館都會就近選擇一個倉庫，使得
最小；
思路：分析一下， 用 dp[i][j] 表示前j家餐館修建i個倉庫的值最小，那么當修建第i個倉庫的時候，前i-1個倉庫肯定修好了，所以用dp[i-1][k]來表示前k家餐館修建i-1個倉庫的最小值，那么容易知道 i-1<=k<j;
所以得到狀態轉移方程：
dp[i][j]=min(dp[i-1][k]+cost[k+1][j]);
PS:cost[i][j]表示在第i家餐館到第j家餐館修建一個倉庫的最小值，那么肯定修建在 (i+j)/2處才是最優的；所以預處理一下cost即可；
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <cstdio>
#include <cmath>
#include <algorithm>
const int INF=99999999;
using namespace std;
int dp[33][220];
int a[220];
int cost[220][220];
int main()
{
int n,m,test=1;
while(cin>>n>>m)
{
if(n==0&&m==0)break;
for(int i=1;i<=n;i++)cin>>a[i];
//預處理一下
memset(cost,0,sizeof(cost));
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++)
{
for(int k=i;k<=j;k++)
cost[i][j]+=fabs(a[k]-a[(i+j)/2]);
}
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)dp[1][i]=cost[1][i];
//dp
for(int i=2;i<=m;i++)
for(int j=i+1;j<=n;j++)
{
dp[i][j]=INF;
for(int k=i-1;k<j;k++)
dp[i][j]=min(dp[i][j],dp[i-1][k]+cost[k+1][j]);
}
printf("Chain %d
",test++);
printf("Total distance sum = %d
",dp[m][n]);
}
return 0;
}
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