首页 > ACM题库 > HDU-杭电 > HDU 1227 Fast Food-动态规划-[解题报告] C++
2013
12-04

HDU 1227 Fast Food-动态规划-[解题报告] C++

Fast Food

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

6 3
5
6
12
19
20
27
0 0

样例输出:

Chain 1
Total distance sum = 8

#include <iostream>
 using namespace std;
 
 int fab(int a)
 {
     return a > 0 ? a : -a;
 }
 
 int cost[210][210];
 int dp[32][210];
 int a[210];
 
 int k, n;
 int main()
 {
     int cas = 0;
     while(scanf("%d%d", &n, &k) && (n || k))
     {
         for(int i = 1; i <= n; ++i)
             scanf("%d", a + i);
         for(int i = 1; i <= n; ++i)
             for(int j = i; j <= n; ++j)
             {
                 cost[i][j] = 0;
                 for(int k = i; k <= j; ++k)
                     cost[i][j] += fab(a[k] - a[(i+j)/2]);
             }
         for(int i = 1; i <= k; ++i)
             for(int j = 1; j <= n; ++j)
                 dp[i][j] = 0;
         for(int i = 1; i <= n; ++i)
             dp[1][i] = cost[1][i];
             
         for(int i = 2; i <= k; ++i)
             for(int j = i + 1; j <= n; ++j)
             {
                 int mmin = 1 << 30;
                 for(int k = i -1; k < j; ++k)
                     if(dp[i-1][k] + cost[k+1][j] < mmin)
                         mmin = dp[i-1][k] + cost[k+1][j];
                 dp[i][j] = mmin;
             }
         printf("Chain %d\n", ++cas);
         printf("Total distance sum = %d\n\n", dp[k][n]);
     }
     return 0;
 }

dp[k][n]

         dp[
i ] [ j ]表示前i个储存站供应前j个快餐店的最短距离。

         dp
[ i ] [ j ]=min { dp[ i-1 ][ k ]+在k到j直接选择一个储存站的最小距离 },k=i+1,i+2,..j-1.

         虽然这些过程中会违反“某个快餐店没有得到距离它最近的存储站的供货”,但是违反了这一性质的必定不是最短的距离,那么也就保持了结果的正确性。

         中位数的性质:求x轴上n个点,那么找一个点到其它点的距离和最小,那么这个点必定是中位数。


  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮