pku1160
DP .首先求出任意两个村庄间(包括端点)建一个邮局后,到邮局的最短距离和,记为min_dis[i][j],当邮局建到中点时min_dis[i][j]最小. 用f[i][j]表示前i个邮局建在前j个村庄.则f[i][j]=min{f[i-1][k]+min_dis[k+1][j]，i-1<=k<j}.见代码：
#include
<
iostream
>
using
namespace
std;
int
v,p;
int
vill[
305
];
int
f[
35
][
305
];
int
min_dis[
305
][
305
];
int
main()
{
int i,j,k; scanf("%d%d",&v,&p); for(i=1;i<=v;i++) scanf("%d",&vill[i]); memset(min_dis,0,sizeof(min_dis)); for(i=1;i<=v;i++) for(j=i+1;j<=v;j++)
{ int mid=(i+j)/2; for(k=i;k<mid;k++) { min_dis[i][j]+=vill[mid]-vill[k];
} for(k=mid+1;k<=j;k++) { min_dis[i][j]+=vill[k]-vill[mid]; } } for(i=1;i<=v;i++) f[1][i]=min_dis[1][i]; for(i=2;i<=p;i++) for(j=i;j<=v;j++) { int minn=100000000; for(k=i-1;k<=j-1;k++) { if(f[i-1][k]+min_dis[k+1][j]<minn) minn=f[i-1][k]+min_dis[k+1][j]; } f[i][j]=minn; } printf("%d\n",f[p][v]); //system("pause"); return 0;
}