P4767 [IOI2000]邮局 - 平行四边形不等式优化DP

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.

在高速公路旁边有一条村庄的直道。高速公路表示为整数轴，并且每个村庄的位置用单个整数坐标标识。同一地点没有两个村庄。两个位置之间的距离是它们整数坐标差的绝对值。

邮局将建在一些但不一定是所有村庄。一个村庄和其中的邮局具有相同的位置。为了建造邮局，应选择其位置，以便每个村庄与其最近的邮局之间的所有距离的总和最小。

你要写一个程序，考虑到村庄的位置和邮局的数量，计算每个村庄和最近的邮局之间所有距离的最小可能总和。

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.

此程序是从标准输入读取。第一行包含两个整数：第一行是村庄的数量V，1 <= V <= 300，第二行是邮局的数量P，1 <= P <= 30，P <= V. 第二行按升序包含V个整数。这V个整数是村庄的位置。对于每个位置X，保证 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.第一行包含一个整数S，它是每个村庄与其最近的邮局之间所有距离的总和。

方法

这道题是一道DP题，由于DP题有一个常见的性质，即 f [ ][ ] 数组中存的子解都具有可输出的共同点，所以这种DP题的状态转移方程都是很好推断的。这道题的 f [ i ] [ j ] 表示前 i 个村庄建 j 个邮局的解，状态转移方程是

$f[i][j]=min(\sum_{k = 1}^{k = i - 1}f[k][j - 1] + w[k + 1][i])$

由于此复杂度是O（ $mn^{2}$ ），但是数据量很小，所以用朴素DP是可以做出来的。

但是，为了提升自我增长知识，我决定优化一下，把复杂度降为O（ $2mn$ ），具体是用一种叫“平行四边形不等式优化DP”的方式，把 K 的枚举平摊成一次O（n）。

判断是否能用此法，主要靠

1）暴力打表找规律

2） 凭直觉

平行四边形优化主要有一个特点，每一个 f [ i ][ j ] 最后的 k 值都大于 f [ i ][ j - 1 ] 的 k ，且小于 f [ i + 1 ][ j ] 的 k 。意思是，状态转移方程可以优化：

$f[i][j]=min(\sum_{k = k_{i,j-1}}^{k = k_{i + 1,j}}f[k][j - 1] + w[k + 1][i])$

主要是因为 w [][] 具有平行四边形不等式的性质，即对于 a < b < c < d ，

w [ a ] [ c ] + w [ b ] [ d ] < w [ a ] [ d ] + w [ b ] [ c ] 。

这可以算是最难的一种方法，因为坑很多，很难AC，但是一般只要样例过了，就离成功不远了。

详见大神题解：https://blog.csdn.net/NOIAu/article/details/72514812

代码：

#include<cstdio>

#include<cstring>

#include<iostream>

using namespace std;

int read() {

    int f = 1,x = 0;char s = getchar();

    while(s < '0' || s > '9') {if(s == '-')f = -1;s = getchar();}

    while(s >= '0' && s <= '9') {x = x * 10 + s - '0';s = getchar();}

    return x * f;

}

int n,m,i,j,k,o;

int pos[3005],f[3005][305],w[3005][3005],sum1[3005],sum2[3005],s[3005][3005];

int main() {

	n = read();m = read();

	for(i = 1;i <= n;i ++) {

		pos[i] = read();

	}

	for(i = 1;i <= n;i ++) {

		sum1[i] = sum1[i - 1] + pos[i] - pos[1];

	}

	for(i = n;i > 0;i --) {

		sum2[i] = sum2[i + 1] + pos[n] - pos[i];

	}

	for(i = 1;i <= n;i ++) {

		for(j = i;j <= n;j ++) {

			if(i == j) w[i][j] = 0;

			else {

				int k = (i + j) / 2;

				w[i][j] = sum1[j] - sum1[k] - (j - k) * (pos[k] - pos[1]) + sum2[i] - sum2[k] - (k - i) * (pos[n] - pos[k]);

			}

		}

	}

	memset(f,0x3f,sizeof(f));

	for(i = 1;i <= n;i ++) {

		f[i][1] = w[1][i];

	}

	f[0][0] = f[1][1] = 0;

	for(i = 1;i <= n;i ++) {

		for(j = min(i,m);j > 0;j --) {

			if(!s[i - 1][j]) s[i - 1][j] = min(i - 1,j - 1);

			if(!s[i][j + 1]) s[i][j + 1] = i - 1;

			for(k = s[i - 1][j];k <= s[i][j + 1];k ++) {

				if(f[k][j - 1] + w[k + 1][i] < f[i][j]) {

					f[i][j] = f[k][j - 1] + w[k + 1][i];

					s[i][j] = k;

				}

			}

		}

	}

	printf("%d",f[n][m]);

    return 0;

}

巴特西

P4767 [IOI2000]邮局 - 平行四边形不等式优化DP

Input

Output

方法

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