pku 1160 Post Office

#include <iostream>#include <cmath>#include <climits>using namespace std;int pos[301];int v[31][301];int cost[301][301];//状态转移方程为: //前i个邮局控制前j个村庄的最小距离和://v[i][j] = min{v[i-1][k]+cost[k+1][j]}(i-1 <= k <= j-1);int main(){int V, P;scanf("%d%d", &V, &P);for(int i = 1; i <= V; ++i)scanf("%d", &pos[i]);//从村落i到村落j由一个邮局控制时的最短距离和for(int i = 1; i <= V; ++i)for(int j = i; j <= V; ++j){int mid = (i+j) >> 1;for(int k = i; k <= j; ++k)cost[i][j] += abs(pos[mid] - pos[k]);}//initialfor(int i = 1; i <= V; ++i)v[1][i] = cost[1][i];//dpfor(int i = 2; i <= P; ++i)for(int j = i; j <= V; ++j){int min = INT_MAX;for(int k = i-1; k <= j-1; ++k)if(v[i-1][k] + cost[k+1][j] < min)min = v[i-1][k] + cost[k+1][j];v[i][j] = min;}printf("%d/n", v[P][V]);return 0;}