【POJ1160】【四边形优化DP】Post Office

最新推荐文章于 2021-04-13 23:04:24 发布

MaverickFW

最新推荐文章于 2021-04-13 23:04:24 发布

阅读量555

点赞数 1

CC 4.0 BY-SA版权

分类专栏： DP POJ 文章标签： dp poj 四边形优化

本文链接：https://blog.youkuaiyun.com/MaverickFW/article/details/72566338

DP 同时被 2 个专栏收录

31 篇文章

订阅专栏

POJ

7 篇文章

订阅专栏

本文介绍了一种使用四边形优化动态规划方法解决邮局选址问题的策略。给定一条直线上有若干村庄，目标是确定最少数量的邮局位置，使得每个村庄到最近邮局的距离之和最小。输入包含村庄数量和邮局数量，以及村庄的位置坐标。输出为最小总距离。例如，对于10个村庄和5个邮局的情况，可以得到总距离为9的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Post Office

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

#include <cstdio>
#include <iostream>
#include <cstring>
#define INF 0x3f

using namespace std;

const int maxn = 305;
int n,m,tmp;
int a[maxn],w[maxn][maxn],s[maxn][maxn],f[maxn][maxn];

template <class T> inline void read(T &x) {
    int flag = 1; x = 0;
    char ch = getchar();
    while(ch <  '0' || ch >  '9') { if(ch == '-')  flag = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = (x<<1)+(x<<3)+ch-'0'; ch = getchar(); }
    x *= flag;
}

int main() {
    read(n); read(m);
    for(int i = 1; i <= n; i++) read(a[i]);
    for(int i = 1; i <= n; i++)
        for(int j = i+1; j <= n; j++) w[i][j] = w[i][j-1]+a[j]-a[(i+j)>>1];
    memset(f, INF, sizeof f);
    for(int i = 1; i <= n; i++) f[i][1] = w[1][i];
    for(int j = 1; j <= m; j++) {
        s[n+1][j] = n;
        for(int i = n; i > j; i--)
            for(int k = s[i][j-1]; k <= s[i+1][j]; k++) {
                tmp = f[k][j-1]+w[k+1][i];
                if(tmp < f[i][j]) f[i][j] = tmp, s[i][j] = k;
            }
    }
    printf("%d\n",f[n][m]);
    return 0;
}