HDU_1227_Fast Food_动态规划

链接：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): 2695 Accepted Submission(s): 1142


[align=left]Problem Description[/align]
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

#include<iostream>
#include<cstdio>
#include<stdlib.h>
#include<cstring>
using namespace std;
#define LL long long

int dis[205];
int dist[205][205];
int dp[35][205];

int main()
{
int n,k,cases=1;
while(scanf("%d%d",&n,&k)!=EOF&&n&&k)
{
memset(dist,0,sizeof(dist));

for(int i=0; i<=k; i++)
for(int j=0; j<=n; j++)
dp[i][j]=2000000000;
//cout<<dp[0][0];
dp[0][0]=0;
dp[1][1]=0;
for(int i=1; i<=n; i++)
scanf("%d",&dis[i]);
for(int i=1; i<=n-1; i++)
for(int j=i; j<=n; j++)
{
int mid=(i+j)/2;
for(int k=i; k<=j; k++)
dist[i][j]+=abs(dis[mid]-dis[k]);
}
for(int i=1;i<=n;i++)
dp[1][i]=dist[1][i];
for(int j=2; j<=n; j++)
for(int i=2; i<=j,i<=k; i++)
for(int m=i-1; m<=j-1; m++)
dp[i][j]=min(dp[i][j],dp[i-1][m]+dist[m+1][j]);

printf("Chain %d\n",cases++);
printf("Total distance sum = %d\n\n",dp[k]
);
}
return 0;
}

View Code