HDU1227 动态规划 四边形不等式优化

Fast Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 740    Accepted Submission(s): 338

Problem Description

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.

 

 

Input

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.

 

 

Output

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.

 

 

Sample Input

6 3

5

6

12

19

20

27

0 0

 

 

Sample Output

Chain 1

Total distance sum = 8

 

 

设 s[i][j] 为第 i 个站到第 j 个站开一家店的最小距离和 , 可以证明 , 在 i 和 j 之间的终点开店的距离和是最小的 .dp[i][j] 为前 i 个站开 j 家店的最小距离和 , 转移方程是 dp[i][j]=min(dp[k][j-1]+s[k+1][i]), 另外 , 这题可以用四边形不等式进行优化 .

 

#include<stdio.h> #include<string.h> #define N 205 #define M 35 #define inf 2000000000 int dp[N][M],w[N][N],a[N],s[N][M]; int main() { int m,n,t,i,j,k,last,cas=0; while (scanf("%d%d",&n,&m)) { cas++; if (n==0 && m==0) break; memset(dp,0,sizeof(dp)); memset(s,0,sizeof(s)); memset(w,0,sizeof(w)); for (i=1;i<=n;++i) scanf("%d",&a[i]); for (j=0;j<n;++j) { for (i=1;i<=n-j;++i) { t=i+j; if (j==0) w[i][t]=0; else if (j%2==0) { w[i][t]=w[i][t-1]+(a[t]-a[(i+t-1)/2+1]); } else { w[i][t]=w[i+1][t]+(a[(i+1+t)/2]-a[i]); } } } for (i=1;i<=n;++i) dp[i][0]=w[1][i]; for (i=1;i<=n;++i) { last=0; for (j=1;j<=(m<i?m:i);++j) { if (j==1) {dp[i][j]=w[1][i];continue;} dp[i][j]=inf; for (k=s[i][j-1];k<i;++k) { if (dp[k][j-1]+w[k+1][i]<dp[i][j]) { dp[i][j]=dp[k][j-1]+w[k+1][i]; s[i][j]=k; } } } } printf("Chain %d/n",cas); printf("Total distance sum = %d/n/n",dp[n][m]); } return 0; }