描述：状态方程p[i][j]=dp[i-1][k]+dist(k+1,j)，由于没搞懂距离dist是怎么计算的，以为是num[j]-num[k+1]，结果wa了一次，在状态转移的时候，采用一个数组sc记录一下节点的位置
#include <cstdio>
#include <cstring>
#define N 0x7fffffff;
int num[210];
int dp[35][210];
int sc[35][210];
void show(int cur,int pos)
{
if(cur>1) show(cur-1,sc[cur][pos]-1);
printf("Depot %d at restaurant %d serves ",cur,(pos+sc[cur][pos])/2);
if(pos-sc[cur][pos]>0) printf("restaurants %d to %d
",sc[cur][pos],pos);
else printf("restaurant %d
",pos);
}
int dist(int x,int y)
{
int v=0,cur=(x+y)/2,p;
for(int i=x; i<=y; ++i)
{
p=(num[cur]-num[i]);
if(p<0) p=-p;
v+=p;
}
return v;
}
int main()
{
// freopen("in.txt","r",stdin);
int n,m,t=1;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(!n&&!m) break;
for(int i=1; i<=n; ++i) scanf("%d",&num[i]);
for(int i=1; i<=n-m+1; ++i) dp[1][i]=dist(1,i),sc[1][i]=1;
for(int i=2; i<=m; ++i)
for(int j=i; j<=n-m+i; ++j)
{
dp[i][j]=N;
for(int k=i-1; k<j; ++k)
{
int v=dist(k+1,j);
if(dp[i][j]>dp[i-1][k]+v)
dp[i][j]=dp[i-1][k]+v,sc[i][j]=k+1;
}
}
printf("Chain %d
",t++);
show(m,n);
printf("Total distance sum = %d
",dp[m][n]);
}
}