USACO 2017 January Contest, Platinum Problem 3. Subsequence Reversal

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USACO2017-JAN-P3

答案:

import java.util.Arrays;
import java.util.Scanner;

public class subrev {
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
a = new int[n];
for (int i = 0; i < n; i++) a[i] = in.nextInt();
dp = new int[n][n][51][51];
for (int[][][] x : dp) for (int[][] y : x) for (int[] z : y) Arrays.fill(z, -1);
System.out.println(solve(0,n-1,0,50));
}

public static int[] a;
public static int[][][][] dp;
public static int solve(int i, int j, int k, int m) {
if (dp[i][j][k][m] != -1) return dp[i][j][k][m];
if (k > m) return -(1 << 29); // impossible case
if (i > j) return 0;
if (i == j) {
if (a[i] >= k && a[i] <= m) return 1;
else return 0;
}
// max value from interval [i,j] given we've taken k from prefix and m from suffix
int res = 0;
// case 1: swap i and j
// case 1a: don't include a[i] or a[j]
res = Math.max(res, solve(i+1,j-1,k,m));
// case 1b: include a[j]
if (a[j] >= k)
res = Math.max(res, solve(i+1,j-1,a[j],m)+1);
// case 1c: include a[i]
if (a[i] <= m)
res = Math.max(res, solve(i+1,j-1,k,a[i])+1);
// case 1d: include both
if (k <= a[j] && a[j] <= a[i] && a[i] <= m) {
res = Math.max(res, solve(i+1,j-1,a[j],a[i])+2);
}

// case 2: don't swap, move i forward
// case 2a: don't include a[i]
res = Math.max(res, solve(i+1,j,k,m));
// case 2b: include a[i]
if (a[i] >= k)
res = Math.max(res, solve(i+1,j,a[i],m)+1);

// case 3: don't swap, move j backward
// case 3a: don't include a[j]
res = Math.max(res, solve(i,j-1,k,m));
// case 3b: include a[j]
if (a[j] <= m)
res = Math.max(res, solve(i,j-1,k,a[j])+1);

return dp[i][j][k][m] = res;
}
}

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