小白dp uva 662 - Fast Food （除夕夜的博客 ^ ^）

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662 - Fast Food

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 of n restaurants along the highway as n integers $d_1 < d_2 < \dots < d_n$ (these are the distances measured from the company's headquarter, which happens to be at the same highway). Furthermore, a number $k (k \leŸ n)$ 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

$\begin{displaymath}\sum_{i=1}^n \mid d_i - (\mbox{position of depot serving restaurant }i) \mid\end{displaymath}$

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.

The input file contains several descriptions of fastfood chains. Each description starts with a line containing the two integers n and k. n and kwill satisfy $1 \leŸ n\leŸ 200$ , $1 \leŸ k Ÿ\le 30$ , $k \le n$ . Following this will n lines containing one integer each, giving the positions d_i of the restaurants, ordered increasingly.

The input file will end with a case starting with n = k = 0. This case should not be processed.

Input

For 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

Output a blank line after each test case.

Sample Input

6 356121920270 0

Sample Output

Chain 1

Depot 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

题意：

给出n个点，要建k个depots，使得每个fastfood离最近的depots的总和最小。

思路：

先预处理sum数组，sum[i][j]-第i个位置到第j个位置建一个depots产生的距离和（建在中点位置距离和最小），预处理也是动态规划递推，画图很好找方程。

然后dp[i][j]表示处理到第i个位置建了j个depots的距离和。

那么有方程：

dp[i][j]=dp[k][j-1]+sum[k+1][i]; （k<i）

用数组存下最佳路径，于是问题解决了。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005#define MAXN 100005#define mod 1000000000#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int n,m,ans,cnt,tot,flag;int a[maxn],res[maxn];int dp[205][35],sum[205][205],pre[205][35];void solve(){    int i,j,t;    memset(dp,0x3f,sizeof(dp));    for(i=1; i<=n; i++) dp[i][1]=sum[1][i];    for(i=1; i<=n; i++)    {        for(j=1; j<=m&&j<=i; j++)        {            for(int k=1; k<i; k++)            {                if(dp[i][j]>dp[k][j-1]+sum[k+1][i])                {                    dp[i][j]=dp[k][j-1]+sum[k+1][i];                    pre[i][j]=k;                }            }        }    }}int main(){    int i,j,t,test=0;    while(scanf("%d%d",&n,&m),n|m)    {        for(i=1; i<=n; i++)        {            scanf("%d",&a[i]);        }        for(i=1; i<=n; i++) sum[i][i]=0;        for(i=1; i<=n; i++)        {            for(j=i+1; j<=n; j++)            {                sum[i][j]=sum[i][j-1]+a[j]-a[(i+j)>>1];            }        }        solve();        printf("Chain %d\n",++test);        t=n;        for(i=m;i>=1;i--)        {            res[i]=t;            t=pre[t][i];        }        res[0]=0;        for(i=1;i<=m;i++)        {            if(res[i-1]+1<res[i]) printf("Depot %d at restaurant %d serves restaurants %d to %d\n",i,(res[i]+res[i-1]+1)>>1,res[i-1]+1,res[i]);            else printf("Depot %d at restaurant %d serves restaurant %d\n",i,res[i],res[i]);        }        printf("Total distance sum = %d\n\n",dp[n][m]);    }    return 0;}

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