百练162:Post Office

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

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. 

输入

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.

输出

The first line contains one integer S, which is the sum of all distances between each village and its nearest post office.

样例输入

10 51 2 3 6 7 9 11 22 44 50

样例输出

9

分析：题目大意是说要在一维坐标轴上的V个村庄建立P个邮局，使村庄到最近邮局的总距离最小，区间动态规划中经典的邮局问题，用dp[i][j]表示前i个村庄建j个邮局，首先求出只建一个邮局的情况，然后再动态计算P（P>=2）个邮局时的最小距离

代码：

#include<cstdio>#define min(a,b) ((a)<(b)?(a):(b))int v,p,a[305],dp[305][305],dis[305][305];int main(){scanf("%d%d",&v,&p);for(int i=1;i<=v;++i) scanf("%d",&a[i]);for(int i=1;i<=v;i++){for(int j=i+1;j<=v;j++)dis[i][j]=dis[i][j-1]+a[j]-a[(i+j)>>1];}for(int i=1;i<=v;++i){dp[i][1] = dis[1][i];}for(int r=2;r<=p;++r){for(int i=r+1;i<=v;++i){dp[i][r]=0x3f3f3f3f;for(int j=r-1;j<i;++j)dp[i][r]=min(dp[i][r],dp[j][r-1]+dis[j+1][i]);}}printf("%d\n",dp[v][p]);return 0;}