1160 Post Office //没经过四边形优化的

Post Office

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10035		Accepted: 5397

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

9

Source

IOI 2000

/* 设m[v, p]为在前v个村庄中修建前p个邮局时的最小距离。按照刚开始的0-1背包的思路可以得到： m[v, p]=min{m[v-1, p], m[v-1, p-1]+dis_cost} 公式很好理解，就是不好计算。 因此，我们可以根据上面的公式初始化m[v, p]矩阵。 再考虑p>1的情况，如果已知m[v, p-1]，那么在v个村庄中已经建了p-1个邮局，如果再建一个邮局，就是求m[v, p]， 这个要好求很多。 考虑在v个村庄中再建一个邮局，就是遍历r~v，（r=p-1），找到一个最小值。因此得到下面的递推关系式： m[v, p] = min{m[i, p-1] + dis(i+1, v), i=p-1, ... v-1} */ #include <cstdio> #include <cstring> #define min(a,b) a>b?b:a; int dp[31][301],sum[301][301]; int p[301]; int n,v; int main() { scanf("%d%d",&n,&v); for(int i=1;i<=n;i++) scanf("%d",&p[i]); for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) { sum[i][j]=sum[i][j-1]+p[j]-p[(i+j) / 2];//可以总结出来的 if(i==1) dp[1][j]=sum[i][j]; } for(int i=2;i<=v;i++) for(int j=i+1;j<=n ;j++) { dp[i][j]=0x7fffffff; for(int k=i-1;k<=j-1;k++) { dp[i][j]=min(dp[i][j],dp[i-1][k]+sum[k+1][j]); } } printf("%d/n",dp[v][n]); return 0; }