NEFU 698 Post office 大概是dp?

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本文介绍了一个经典的邮局选址问题：在一维直线上寻找最优位置建立邮局，使得该邮局到指定范围内各村庄的距离之和最小。文章提供了一种通过预处理的方法来高效求解此问题，并附带了完整的C++代码实现。

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Post office

Time Limit 1000ms

Memory Limit 65536K

description

There are N(N<=1000) villages along a straight road, numbered from 1 to N for simplicity. We know exactly the position of every one (noted pos[i],pos[i] is positive integer and pos[i]<=10^8). The local authority wants to build a post office for the people living in the range i to j(inclusive). He wants to make the sum of |pos[k]-position_of_postoffice| (i<=k<=j) is minimum.

input

　　For each test case, the first line is n. Then n integer, representing the position of every village and in acending order. Then a integer q (q<=200000), representing the queries. Following q lines, every line consists of two integers i and j. the input file is end with EOF. Total number of test case is no more than 10.
　　Be careful, the position of two villages may be the same.

output

　　For every query of each test case, you tell the minimum sum.

sample_input

sample_output

2
1

hint

　　Huge input,"scanf" is recommend.

------------------

可以用dp进行一下预处理=。=

但是！！！！暴力也能过！！！！！

------------------

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

long long a[1111];
long long f[1111];
long long g[1111];
int n;
int q;


int main()
{
    while (~scanf("%d",&n))
    {
        if (n==0) break;
        memset(a,0,sizeof(a));
        memset(f,0,sizeof(f));
        memset(g,0,sizeof(g));

        for (int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
            f[i]=f[i-1]+a[i];
        }
        for (int i=n;i>=1;i--)
        {
            g[i]=g[i+1]+a[i];
        }
        scanf("%d",&q);
        while (q--)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            int mid=(x+y)/2;
            long long ans=(a[mid]*(mid-x)-(f[mid-1]-f[x-1]))+((g[mid+1]-g[y+1])-a[mid]*(y-mid));
            printf("%I64d\n",ans);
        }
    }
    return 0;
}