poj 1160 Post Office

最新推荐文章于 2021-04-13 23:04:24 发布

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最新推荐文章于 2021-04-13 23:04:24 发布

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本文深入探讨了AI音视频处理领域中的关键技术——视频分割与语义识别，通过详细解释这些技术的工作原理、应用场景及实际应用案例，旨在帮助开发者更好地理解和应用这些技术。视频分割技术能够将视频内容高效地划分成多个部分，而语义识别则能理解视频中的元素及其意义，为后续的智能分析和决策提供基础。

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Post Office

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13983		Accepted: 7540

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

Source

IOI 2000

考的是基础数学知识呀。
可惜基础没学好搞了半天。
dp[i][j]=min(dp[i][j],dp[k][j-1]+dis[k+1][i]);
k为1到i-1中的数。看方程应该可以看懂吧。
dis[i][j]=dis[i][j-1]+pos[j]-pos[(i+j)/2];

#include <iostream>
#include<string.h>
#include<stdio.h>
#define MIN(a,b) ((a)<(b)?(a):(b))
using namespace std;

int dis[310][310];//dis[i][j]记录在i和j之间建一个邮局的最小花费
int dp[310][310];//dp[i][j]记录在前i个村子建j个邮局且i村庄由前面的邮局管辖的最小花费
int pos[310];//记录村子下标
int main()
{
    int i,j,k,v,p;

    while(~scanf("%d%d",&v,&p))
    {
        for(i=1; i<=v; i++)
            scanf("%d",&pos[i]);
        memset(dp,0x3f,sizeof dp);
        memset(dis,0,sizeof dis);
        for(i=1; i<=v; i++)
        {
            for(j=i+1; j<=v; j++)
                dis[i][j]=dis[i][j-1]+pos[j]-pos[(i+j)/2];
            dp[i][1]=dis[1][i];
            dp[i][i]=0;
            //printf("dp[%d][1] %d\n",i,dp[i][1]);
            //getchar();
        }
        for(j=2; j<=p; j++)//循环必须这么写。这样才能保证要利用的前面的结果已算出
            for(i=j+1; i<=v; i++)
                for(k=1; k<i; k++)
                    dp[i][j]=MIN(dp[i][j],dp[k][j-1]+dis[k+1][i]);
        printf("%d\n",dp[v][p]);
    }
    return 0;
}