Uva10271-Chopsticks (recursive)

Uva10271-Chopsticks (recursive)

Question: uva10271-Chopsticks (recursive)

N pairs of chopsticks (A, B, C) and badness (B-A) must be selected from these chopsticks) ^ 2. minimum badness is required.

Solution: Select adjacent chopsticks for A and B. Such badness must be relatively small. However, it is troublesome to consider C. Finally, I read the question of the great god. The chopsticks should start from the length to the short. dp [k] [j]: The first j chopsticks have K pairs, and the smallest badness. Dp [k] [j] = Min (dp [k] [j-1], dp [k-1] [j-2] + badness (j-1, j ). check whether j chopsticks are added to the k pair. If j is added, j must be paired with j-1, so that badness is relatively small. Then, because it is from the largest to the smallest, the front must be longer than the back. Therefore, you only need to check whether j is greater than or equal to 3 * k to eliminate the interference of C.

Code:

#include 
 
  #include 
  
   typedef long long ll;const int N = 5005;const int M = 1010;const ll INF = 1e13;ll dp[M][N];int c[N];ll Min (const ll a, const ll b) { return a < b? a: b; }int W (const int a, const int b ) { return (a - b) * (a - b); }int main () {int t;int k, n;scanf ("%d", &t);while (t--) {scanf ("%d%d", &k, &n);for (int i = n; i >= 1; i--)scanf ("%d", &c[i]);k += 8;for (int i = 1; i <= n; i++)dp[0][i] = 0;for (int i = 1; i <= k; i++)for (int j = 1; j <= n; j++) {dp[i][j] = INF;if (i * 3 > j)continue;if (j - 2 >= 1)dp[i][j] = Min (dp[i][j], dp[i - 1][j - 2] + W(c[j], c[j - 1]));if (j - 1 >= 1)dp[i][j] = Min (dp[i][j], dp[i][j - 1]);}ll ans = INF;for (int i = 3 * k; i <= n; i++)ans = Min (ans, dp[k][i]);printf ("%lld\n", ans);}return 0;}