hdu 2639 Bone Collector II【01背包第k优解】

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

Problem Description

The title of this problem is familiar,isn’t it?yeah,if you had took part in the “Rookie Cup” competition,you must have seem this title.If you haven’t seen it before,it doesn’t matter,I will give you a link:

Here is the link:http://acm.hdu.edu.cn/showproblem.php?pid=2602

Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum.

If the total number of different values is less than K,just ouput 0.

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the K-th maximum of the total value (this number will be less than 231).

Sample Input

3
5 10 2
1 2 3 4 5
5 4 3 2 1
5 10 12
1 2 3 4 5
5 4 3 2 1
5 10 16
1 2 3 4 5
5 4 3 2 1

Sample Output

12
2
0

背包九讲中写到，第k优解的基本思想是，将每个状态都表示成有序队列，将状态转移方程中的max/min转化成有序队列的合并。
这个题时个第k优解的裸题
代码：

#include 
using namespace std;
int dp[1005][35], c[105], w[105], a[35], b[35];
int main() {
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while (T--) {
        int n, v, k;
        cin >> n >> v >> k;
        for (int i = 1; i <= n; i++) cin >> w[i];
        for (int i = 1; i <= n; i++) cin >> c[i];
        memset(dp, 0, sizeof dp);
        for (int i = 1; i <= n; i++) {
            for (int j = v; j >= c[i]; j--) {
                for (int p = 1; p <= k; p++) {
                    a[p] = dp[j][p];
                    b[p] = dp[j - c[i]][p] + w[i];
                }
                a[k + 1] = -1, b[k + 1] = -1;
                int x = 1, y = 1, z = 1;
                while (z <= k && (a[x] != -1 || b[y] != -1)) {
                    if (a[x] > b[y]) dp[j][z] = a[x++];
                    else dp[j][z] = b[y++];
                    //要根据题意是否把不同策略得到的解当作一个解
                    if (dp[j][z] != dp[j][z - 1]) z++;//除去重复解
                }
            }
        }
        cout << dp[v][k] << endl;
    }
    return 0;
}