zoj 1196 Fast Food 变种区间dp
2016-05-10 10:01
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好吧,其实我觉得这个题在一开始想的时候与区间dp还是有点差别的……虽然写出来跟区间dp确实差不多
Fast Food
Time Limit: 2 Seconds
Memory Limit: 65536 KB
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
![](http://acm.zju.edu.cn/onlinejudge/showImage.do?name=0000%2F1196%2F1196.gif)
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
题意:在一条路上有n个餐馆,n个餐馆的位置给出,现在让你建k个仓库,每个餐馆会由最近的仓库进货,进货的代价为pos餐馆-pos仓库。现在让你计算最小的总进货代价
思路:一开始想到的是用区间合并的方法,这个问题相当于将n个递增的数划分进k个集合,在每个集合里可以很容易想到在序号最中间的餐馆建仓库是最优的,假设这个区间是(i,j)那么在(i+j)/2是最优的,如果往左移动x个单位,那么所有右边的餐馆都加上x个单位,但是左边减去的代价是小于等于x*左边餐馆数的,往右同理。所以在(i+j)/2位置建仓库一定是代价最小的一种情况。但是如果枚举区间的话,最多有c(199,29)种情况,这肯定是超时的。
这时就想到用dp优化了,设dp[i][j]为前i个餐馆建j个仓库的最小代价,那么转移方程就可以写为dp[i][j]=min{dp[i][j],dp[k][j-1]+dis[k+1][i],k=j-1,j……i-1},(dis[i][j]表示在i,j间建立仓库所需的最小代价),其实跟区间dp的思想一样,枚举出分割餐馆的k,前一部分用已求出的dp[k][j-1]得出,后一部分再用dis数组得出。至于求dis数组,只要200*200*200的复杂度,是不会超时的。
代码如下:
dp是世界上最好的算法!!!
Fast Food
Time Limit: 2 Seconds
Memory Limit: 65536 KB
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
![](http://acm.zju.edu.cn/onlinejudge/showImage.do?name=0000%2F1196%2F1196.gif)
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
题意:在一条路上有n个餐馆,n个餐馆的位置给出,现在让你建k个仓库,每个餐馆会由最近的仓库进货,进货的代价为pos餐馆-pos仓库。现在让你计算最小的总进货代价
思路:一开始想到的是用区间合并的方法,这个问题相当于将n个递增的数划分进k个集合,在每个集合里可以很容易想到在序号最中间的餐馆建仓库是最优的,假设这个区间是(i,j)那么在(i+j)/2是最优的,如果往左移动x个单位,那么所有右边的餐馆都加上x个单位,但是左边减去的代价是小于等于x*左边餐馆数的,往右同理。所以在(i+j)/2位置建仓库一定是代价最小的一种情况。但是如果枚举区间的话,最多有c(199,29)种情况,这肯定是超时的。
这时就想到用dp优化了,设dp[i][j]为前i个餐馆建j个仓库的最小代价,那么转移方程就可以写为dp[i][j]=min{dp[i][j],dp[k][j-1]+dis[k+1][i],k=j-1,j……i-1},(dis[i][j]表示在i,j间建立仓库所需的最小代价),其实跟区间dp的思想一样,枚举出分割餐馆的k,前一部分用已求出的dp[k][j-1]得出,后一部分再用dis数组得出。至于求dis数组,只要200*200*200的复杂度,是不会超时的。
代码如下:
#include<stdio.h> #include<string.h> #include<algorithm> int dp[202][31],dis[202][202],map[202]; int main() { int i,j,k,m,n,t; t=0; while(~scanf("%d%d",&n,&k)) { memset(dp,0,sizeof(dp)); memset(dis,0,sizeof(dis)); if(n==0 && k==0)break; t++; printf("Chain %d\n",t); for(i=1;i<=n;i++) scanf("%d",&map[i]); for(i=1;i<=n;i++) for(j=1;j<=n;j++) { int sum=0; int temp=(i+j)/2; for(int i1=i;i1<temp;i1++)sum+=(map[temp]-map[i1]); for(int i1=temp+1;i1<=j;i1++)sum+=map[i1]-map[temp]; dis[i][j]=sum; } for(i=1;i<=n;i++)dp[i][0]=0x1f1f1f; for(i=1;i<=n;i++) for(j=1;j<=i && j<=k;j++) { dp[i][j]=0x1f1f1f; for(int k1=j-1;k1<=i-1;k1++) if(dp[k1][j-1]+dis[k1+1][i]<dp[i][j]) dp[i][j]=dp[k1][j-1]+dis[k1+1][i]; } printf("Total distance sum = %d\n\n",dp [k]); } return 0; }
dp是世界上最好的算法!!!
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