UVA 10271--chopsticks +DP

The length of the chopsticks from long to short arrangement, you can ignore the impact of Z (for each pair (x, y) can always be guaranteed to be greater than its z);

Then define the state Dp[i][j] to represent the minimum cost of the J pair in the front I chopsticks, and then:

If J<3*i is dp[i][j]=dp[i-1][j]

If the j>=3*i for the first chopsticks can be selected to form the J pair (and i-1 chopsticks constitute the J (x, Y)), of course, can also be selected;

Then Dp[i][j]=min (Dp[i-1][j],dp[i-2][j]+w (a[i],a[i-1]))

The code is as follows:

#include <iostream> #include <cstdio> #include <cstring>using namespace std; #define INF 0x3fffffffint A[10000];int dp[5500][1100];int n,k;void input () {    scanf ("%d%d", &k,&n);    for (int i=n;i>=1;i--)        scanf ("%d", &a[i]);} void Solve () {for    (int. i=0;i<=n;i++) for      (int j=0;j<=n;j++)      {          dp[i][j]=inf;          if (j==0)            dp[i][0]=0;      }    for (int i=1;i<=n;i++) for     (int j=1;j<=k+8;j++)     {         dp[i][j]=dp[i-1][j];         if (i>=j*3)         {             dp[i][j]=min (dp[i][j],dp[i-2][j-1]+ (a[i]-a[i-1]) * (a[i]-a[i-1]));         }     }    printf ("%d\n", Dp[n][k+8]);} int main () {    int t;    scanf ("%d", &t);    while (t--)    {       input ();       Solve ();    }  return 0;} </span>

UVA 10271--chopsticks +DP