01背包之求第K优解——Bone Collector II

http://acm.hdu.edu.cn/showproblem.php?pid=2639

题目大意是，往背包里赛骨头，求第K优解，在普通01背包的基础上，增加一维空间，那么F[i,v,k]可以理解为前i个物品，放入容量v的背包时，第K优解的值。时间复杂度为O(NVK)。

Talk is cheap.

看代码吧。

import java.util.Scanner;
 
public class BoneCollector {
    public static void main(String[] sure) {
        int t;
        Scanner sc = new Scanner(System.in);
        t = sc.nextInt();
        while (t-- > 0) {
            int n, v, k;
            n = sc.nextInt();
            v = sc.nextInt();
            k = sc.nextInt();
            int[] val = new int[n + 1];
            int[] vol = new int[n + 1];
            int[][] dp = new int[v + 2][k + 2];
            int[] tp_a = new int[k + 2];
            int[] tp_b = new int[k + 2];
 
            for (int i = 0; i < n; i++) {
                val[i] = sc.nextInt();
            }
            for (int i = 0; i < n; i++) {
                vol[i] = sc.nextInt();
            }
            for (int i = 0; i < n; i++) {
                for (int j = v; j >= vol[i]; j--) {
                    for (int m = 1; m <= k; m++) {
                        tp_a[m] = dp[j][m];
                        tp_b[m] = dp[j - vol[i]][m] + val[i];
                    }
                    int tmp = 1, a = 1, b = 1;
                    tp_a[k+1] = tp_b[k+1] = -1;
                    //循环，依次将前K优的存到dp数组
                    while (tmp <= k && (a <= k || b <= k)) {
                        if (tp_a[a] > tp_b[b]) {
                            dp[j][tmp] = tp_a[a];
                            a++;
                        } else {
                            dp[j][tmp] = tp_b[b];
                            b++;
                        }
                        if (dp[j][tmp] != dp[j][tmp - 1])
                            tmp++;
                    }
                }
            }
            System.out.println(dp[v][k]);
        }
    }
}