Post Office
Time Limit: 1000MS Memory Limit: 10000K
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
题意： 给出n个城市的绝对位置，在某些城市上建立m个邮局，使得所有城市和离它最近的邮局的距离之和最小。
思路： 以前做过这种DP，数学知识：中位数距离其他所有数字的距离之和最短。
证明可以看看这儿：HDU-1227
思路和代码几乎一模一样。
代码如下：
/*
* ID: j.sure.1
* PROG:
* LANG: C++
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <iostream>
#define PB push_back
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
/****************************************/
const int N = 305, M = 35;
int n, m;
int w[N];
int dp[M][N], cost[N][N];
void pre_treat()
{
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
cost[i][j] = 0;
int mid = (i+j)>>1;
for(int k = i; k <= j;k++) {
cost[i][j] += abs(w[mid] - w[k]);
}
}
}
}
int solve()
{
for(int j = 1; j <= n; j++) {
dp[1][j] = cost[1][j];
}
for(int i = 2; i <= m; i++) {
for(int j = 1; j <= n; j++) {
int ret = INF;
for(int k = 1; k < j; k++) {
ret = min(ret, dp[i-1][k] + cost[k+1][j]);
}
dp[i][j] = ret;
}
}
return dp[m][n];
}
int main()
{
#ifdef J_Sure
freopen("000.in", "r", stdin);
//freopen("999.out", "w", stdout);
#endif
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) {
scanf("%d", &w[i]);
}
pre_treat();
int ans = solve();
printf("%d\n", ans);
return 0;
}