Post Office
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 15966 | Accepted: 8671 |
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
思路：
sum[i][j]:在第i个村庄和第j个村庄之间建一个邮局的最短距离；
【1，3】邮局建在第2个村庄；
【1，4】邮局建在2、3距离一样；
【1，5】sum[1][5]=sum[1][4]+村庄5到3的距离；
即：sum[i][j]=sum[i][j-1]+x[j]-x[(i+j)/2];
dp[i][j]:在前i个村庄中建立j个邮局的最小距离；
dp[i][j]=min(dp[i][j],dp[k][j-1]+sum[k+1][i]) ；即在[1,K]村庄建立j-1个邮局，在[k+1,i]村庄建立一个邮局
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define N 305
const int inf=0x3fffffff;
int dp[N][35]; //在前i个村庄中建立j个邮局的最小耗费
int sum[N][N];//sum[i][j]：第i个村庄到第j个村庄建一个邮局的最短距离
int x[N]; //村庄位置
int main()
{
int v,p,i,j,k;
while(scanf("%d%d",&v,&p)!=-1)
{
for(i=1;i<=v;i++)
scanf("%d",&x[i]);
memset(sum,0,sizeof(sum));
for(i=1;i<=v;i++)
{
for(j=i+1;j<=v;j++)
{
sum[i][j]=sum[i][j-1]+x[j]-x[(i+j)/2];
}
}
for(i=1;i<=v;i++)
{
dp[i][i]=0; //一个村庄一个邮局距离为零
dp[i][1]=sum[1][i]; 前i个村庄建立一个邮局
}
for(j=2;j<=p;j++)
{
for(i=j+1;i<=v;i++)
{
dp[i][j]=inf;
for(k=j-1;k<i;k++)
{
dp[i][j]=min(dp[i][j],dp[k][j-1]+sum[k+1][i]);
}
}
}
printf("%d\n",dp[v][p]);
}
return 0;
}