【POJ 1160】Post Office

Post Office
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 16454
 
Accepted: 8915
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
사각형 부등식 최적화 dp.
먼저 소박한 dp방정식을 말한다.
f[i][j]는 전 j개 마을에 i개 우체국을 세운 최소 대가를 나타낸다. 
w[i][j]는 i에서 j 사이에 우체국을 짓는 최소 대가를 표시한다(분명히 중간 대가가 가장 적다)
f[i][j]=min(f[i-1][k]+w[k+1][j])
이렇게 옮기는 것은 O(n^3)의 것이다.
사각형 부등식 최적화 dp조상 논문.
사각형 부등식 최적화의 대략적인 절차는 다음과 같다.
만약 w[i][j]가 사각형의 부등식을 만족시킨다면 w[i][j]+w[i'][j']<=w[i][j']+w[i'][j](i<=i'<=j<=j'),
그러면 f[i][j]도 사각형의 부등식을 만족시키고,
f[i][j]=f[i-1][k]+w[k+1][j], f[i][j]를 설정하면 k에서 최우수치를 얻고 s[i][j]=k를 기록하면 s[i][j]가 단조성을 만족시킨다.
s[i-1][j]<=s[i][j]<=s[i][j+1]
따라서 k는 1부터 매거할 필요가 없다. s[i-1][j]에서 s[i][j+1]까지 매거하면 된다.
여기에 상세한 증명이 있다.
일반적으로 dp방정식이 사각형의 부등식을 만족시키는 것이 비교적 복잡하다는 것을 증명하기 때문에 우리는 s[i][j]폭력을 구하고 s[i-1][j]<=s[i][j]<=s[i][j+1]를 만족시키는지 확인하면 된다.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int w[305][305],f[305][305],s[305][305],n,m,p[305];
void Getw()
{
	for (int i=1;i<=n;i++)
	{
		w[i][i]=0;
		for (int j=i+1;j<=n;j++)
			w[i][j]=w[i][j-1]+p[j]-p[(i+j)>>1];
	}
}
int main()
{
        scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d",&p[i]);
	Getw();
	memset(f,127,sizeof(f));
	for (int i=1;i<=n;i++)
		f[1][i]=w[1][i],s[1][i]=0;
	for (int i=2;i<=m;i++)
	{
		s[i][n+1]=n;
		for (int j=n;j>=i;j--)
			for (int k=s[i-1][j];k<=s[i][j+1];k++)
				if (f[i-1][k]+w[k+1][j]<f[i][j])
					s[i][j]=k,f[i][j]=f[i-1][k]+w[k+1][j];
	}
	printf("%d
",f[m][n]);
	return 0;
}