Post Office - POJ 1160 dp

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本文介绍了一个经典的邮局选址问题，即如何在多个村庄中选择若干地点建立邮局，以使所有村庄到最近邮局的总距离最小。文章提供了一种有效的动态规划解决方案，并附带了完整的AC代码。

Post Office

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15376		Accepted: 8330

Description

There is a straight highway with villages alongside the highway. The highway is represented as an integer axis, and the position of each village is identified with a single integer coordinate. There are no two villages in the same position. The distance between two positions is the absolute value of the difference of their integer coordinates.

Post offices will be built in some, but not necessarily all of the villages. A village and the post office in it have the same position. For building the post offices, their positions should be chosen so that the total sum of all distances between each village and its nearest post office is minimum.

You are to write a program which, given the positions of the villages and the number of post offices, computes the least possible sum of all distances between each village and its nearest post office.

Input

Your program is to read from standard input. The first line contains two integers: the first is the number of villages V, 1 <= V <= 300, and the second is the number of post offices P, 1 <= P <= 30, P <= V. The second line contains V integers in increasing order. These V integers are the positions of the villages. For each position X it holds that 1 <= X <= 10000.

Output

The first line contains one integer S, which is the sum of all distances between each village and its nearest post office.

Sample Input

10 5
1 2 3 6 7 9 11 22 44 50

Sample Output

题意：在v个村庄之间设立p个邮局，使得所有村庄到最近的邮局的距离和最小。

思路：先求出sum[i][j]为i到j的村庄之间如果设立一个邮局的距离最小和，sum[i][j]=sum[i][j-1]+num[j]-num[(i+j)/2];因为邮局设在中间肯定是距离最小的，偶数个的话，中间的两个是一样的。然后遍历k使得dp[i][j]=min(dp[i][j],dp[k][j-1]+sum[k+1][i]);表示到i个村庄设立j个邮局的最短距离。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int sum[310][310],num[310],dp[310][40];
int main()
{ int i,j,k,v,p;
  scanf("%d%d",&v,&p);
  for(i=1;i<=v;i++)
   scanf("%d",&num[i]);
  for(i=1;i<=v;i++)
   for(j=i+1;j<=v;j++)
    sum[i][j]=sum[i][j-1]+num[j]-num[(i+j)/2];
  for(i=1;i<=v;i++)
  { dp[i][1]=sum[1][i];
    for(j=2;j<=p;j++)
    { dp[i][j]=dp[1][j-1]+sum[2][i];
      for(k=2;k<i;k++)
       dp[i][j]=min(dp[i][j],dp[k][j-1]+sum[k+1][i]);
    }
  }
  printf("%d\n",dp[v][p]);
}