小白dp uva 662 - Fast Food （除夕夜的博客 ^ ^）
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 (these are the distances measured from the company's headquarter, which happens to be at the same highway). Furthermore, a number 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.
The input file will end with a case starting with n = k = 0. This case should not be processed.
Input
Output
Output a blank line after each test case.
Sample Input
6 3 5 6 12 19 20 27 0 0
Sample Output
Chain 1
Depot 1 at restaurant 2 serves restaurants 1 to 3 Depot 2 at restaurant 4 serves restaurants 4 to 5 Depot 3 at restaurant 6 serves restaurant 6 Total 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-6
typedef 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;
}