xtu summer individual 5 F - Post Office

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Post OfficeTime Limit: 1000msMemory Limit: 10000KBThis problem will be judged on PKU. Original ID: 1160
64-bit integer IO format: %lld      Java class name: Main  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. 
 InputYour 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. OutputThe first line contains one integer S, which is the sum of all distances between each village and its nearest post office. Sample Input

10 51 2 3 6 7 9 11 22 44 50

Sample Output

9

SourceIOI 2000 解題：dp,dp[i][j]表示i個郵局負責j個村子時的最短距離和。距離和最少？肯定是選取的點如果在第i個與第n-i 個村子的中點(1 <= 1 <= n/2)，距離和會最小啊! 

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 #define INF 0x3f3f3f3f11 using namespace std;12 int dp[35][301],x[301],n,m;13 int dis(int i,int j){14     int sum = 0;15     while(i < j) sum += x[j--]-x[i++];16     return sum;17 }18 int main(){19     int i,j,k;20     while(~scanf("%d%d",&m,&n)){21         memset(dp,0,sizeof(dp));22         for(i = 1; i <= m; i++){23             scanf("%d",x+i);24             dp[1][i] = dis(1,i);25         }26         for(i = 2; i <= n; i++){27             for(j = i; j <= m; j++){28                 dp[i][j] = INF;29                 for(k = i-1; k < j; k++)30                     dp[i][j] = min(dp[i][j],dp[i-1][k]+dis(k+1,j));31             }32         }33         cout<<dp[n][m]<<endl;34     }35     return 0;36 }

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