菜雞選手只能刷刷水題了,這個題O(n^3)的dp還是很顯然的.

#include <bits/stdc++.h>
#define ll long long
using namespace std;
/*To confirm a set of k elements,
  dp is always the best way considering the k elements are in range [0,255].
  So we can set dp[i][j],representing the minimum answer that we have used j elements,
  and we have get the first i initial elements done.
  After pre-processing f[i][j] as the minimum answer for range[i,j] in the array
  in O(n^3) time,we can use dp to solve the problem in O(n^3),too.
*/
ll f[257][257];
ll dp[257][257];
struct node
{ll p,num;
}t[257];
int n,k;
int main (){
int i,j,l;
scanf ("%d%d",&n,&k);
for (i=1;i<=n;i++)
{scanf ("%lld%lld",&t[i].p,&t[i].num);}
memset (dp,0x7f,sizeof(dp));
memset (f,0x7f,sizeof(f));
for (l=0;l<=255;l++)
{for (i=1;i<=n;i++)
{ll num=0;
for (j=i;j<=n;j++)
{num+=((ll)((t[j].p-l)*(t[j].p-l)))*t[j].num;
if (num<f[i][j]) {f[i][j]=num;}
}
}
}
dp[0][0]=0;
for (i=1;i<=n;i++)
{for (j=1;j<=k;j++)
{for (l=0;l<i;l++)
{dp[i][j]=min(dp[i][j],dp[l][j-1]+f[l+1][i]);}
}
}
printf ("%lld\n",dp[n][k]);
return 0;
}