POJ1160 Post Office[序列DP]

Post Office

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18680		Accepted: 10075

Description

There is a straight highway with villages alongside the highway. The highway is represented as an integer axis, and the position of each village is identified with a single integer coordinate. There are no two villages in the same position. The distance between two positions is the absolute value of the difference of their integer coordinates. 

Post offices will be built in some, but not necessarily all of the villages. A village and the post office in it have the same position. For building the post offices, their positions should be chosen so that the total sum of all distances between each village and its nearest post office is minimum. 

You are to write a program which, given the positions of the villages and the number of post offices, computes the least possible sum of all distances between each village and its nearest post office. 

Input

Your program is to read from standard input. The first line contains two integers: the first is the number of villages V, 1 <= V <= 300, and the second is the number of post offices P, 1 <= P <= 30, P <= V. The second line contains V integers in increasing order. These V integers are the positions of the villages. For each position X it holds that 1 <= X <= 10000.

Output

The first line contains one integer S, which is the sum of all distances between each village and its nearest post office. 

Sample Input

10 5
1 2 3 6 7 9 11 22 44 50

Sample Output

9

Source

IOI 2000

－－－－－－－－－－－－－－－－－－－－－

在村庄内建邮局，要使村庄到邮局的距离和最小

－－－－－－－－－－－－－－－－－－－－－

f[i][j]表示前i个村庄中建j个邮局的最小距离,考虑最后一个邮局的位置

f[i][j]=min{f[k][j-1]+l[k+1][i]}

l可以递推，建在中点最好，发现l[i][j]=l[i][j-1]+a[j]-a[(i+j)/2]成立

注意预处理

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int V=305,P=35,INF=1e9;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int v,p,a[V],l[V][V],f[V][P];
void dp(){
    for(int i=1;i<=v;i++)
        for(int j=i+1;j<=v;j++)
            l[i][j]=l[i][j-1]+a[j]-a[(i+j)/2];
    for(int i=1;i<=v;i++) for(int j=0;j<=v;j++) f[i][j]=INF,f[i][1]=l[1][i];
    for(int i=1;i<=v;i++)
        for(int j=1;j<=p;j++){
            for(int k=1;k<i;k++)
                f[i][j]=min(f[i][j],f[k][j-1]+l[k+1][i]);
        }
}
int main(int argc, const char * argv[]) {
    v=read();p=read();
    for(int i=1;i<=v;i++)a[i]=read();
    dp();
    printf("%d",f[v][p]);
    return 0;
}