POJ——T 1160 Post Office

http://poj.org/problem?id=1160

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20218		Accepted: 10899

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

 

 

minx[i][j]表示在i到j个村庄中，放1个邮局的最小距离和、

 　　当 i 到 j 有三个村庄，a1, a2, a3 那么邮局一定选在a2，当有a1,a2,a3,a4时，选在a2,a3是等效的，增加a5时，选在a3，最优，

  　　以此类推，得到 minx[i][j] = minx[i][j-1]+x[j]-x[i+j>>1] ）

f[i][k]表示，从第1个村庄到第i个村庄放了k个邮局的最小距离和、

那么 f[i] [k] =min{ f[j] [k-1]+minx[j+1][i] }(j<i)

 

#include <cstring>
#include <cstdio>

inline void read(int &x)
{
    x=0; register char ch=getchar();
    for(; ch>'9'||ch<'0'; ) ch=getchar();
    for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
}

const int N(323);
#define min(a,b) (a<b?a:b)
int n,m,x[N],minx[N][N],f[N][N];

int Presist()
{
//    freopen("post.in","r",stdin);
//    freopen("post.out","w",stdout);
    
    read(n),read(m);
    if(m==n) { puts("0"); return 0; }
    for(int i=1; i<=n; ++i) read(x[i]);
    for(int i=1; i<=n; ++i)
      for(int j=i+1; j<=n; ++j)
        minx[i][j]=minx[i][j-1]+x[j]-x[i+j>>1];
    memset(f,127/3,sizeof(f));
    for(int i=1; i<=n; ++i)
        f[i][1]=minx[1][i],f[i][i]=0;
    for(int k=2; k<=m; ++k)
      for(int i=1; i<=n; ++i)
      {
          for(int j=i-1; j; --j)
          f[i][k]=min(f[i][k],f[j][k-1]+minx[j+1][i]);
      }
    printf("%d
",f[n][m]);
    return 0;
}

int Aptal=Presist();
int main(){;}