POJ 1160 Post Office(动态规划)

//经典动态规划//dp(i,j)表示用i个邮局，从1到j村庄的最优解//dis(i,j)表示只用1个邮局从i到j村庄的最优解,显然取i,j中点的村庄作为邮局点是最优的//动态转移方程：dp(i,j) = min{dp(i-1,k) + dis(k+1,j)}(i <= k <= j-1)//注意dp数组的初始化#include<iostream>#define INF 1000000000using namespace std;int d[305];int dp[35][305],dis[305][305];int n,m;void init(){memset(dis,0,sizeof(dis));for(int i = 1;i <= n;++i)for(int j = i+1;j <= n;++j){int mid = (i+j) >> 1;for(int k = i;k <= j;++k)dis[i][j] += abs(d[k] - d[mid]);}}void DP(){for(int i = 1;i <= n;++i)dp[1][i] = dis[1][i];for(int i = 2;i <= m;++i)for(int j = 1;j <= n;++j)dp[i][j] = INF;//初始化for(int i = 2;i <= m;i++)for(int j = i;j <= n;++j)for(int k = 1;k <= j;++k){if(dp[i][j] > dp[i-1][k] + dis[k+1][j])dp[i][j] = dp[i-1][k] + dis[k+1][j]; }printf("%d/n",dp[m][n]);}int main(){while(scanf("%d%d",&n,&m) != EOF){for(int i = 1;i <= n;++i)scanf("%d",&d[i]);init();DP();}return 0;}