（01背包 第k优解） Bone Collector II（hdu 2639）

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

 

思路：

对于求最优解的情况，我们对每一种状态只保存了该状态下的最优解，忽略了其他解，进而实现状态之间的转移，而对于求第K优解的情况呢？其实只需要保存每一种状态下的前K优解，从这K个状态进行状态间的转移，同时去重，保存当前状态的K优解即可。（感觉时间复杂度还是挺高的）

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

const int N = 6600;
const int INF = 0x3fffffff;
const long long MOD = 1000000007;
typedef long long LL;
#define met(a,b) (memset(a,b,sizeof(a)))

int dp[N][35];
int a[N], b[N], c[N];
///dp[j][k] 代表容量为 j 的背包的第 k+1 优解

int cmp(int a, int b)
{
    return a > b;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int i, j, k, n, v;

        scanf("%d%d%d", &n, &v, &k);

        met(a, 0);
        met(b, 0);
        met(dp, 0);

        for(i=1; i<=n; i++)
            scanf("%d", &a[i]);
        for(i=1; i<=n; i++)
            scanf("%d", &b[i]);

        for(i=1; i<=n; i++)
        {
            for(j=v; j>=b[i]; j--)
            {
                int w = 0;
                for(int z=0; z<k; z++) ///每次只需考虑前 k 优解的状态转换即可
                {
                    c[w++] = dp[j][z];
                    c[w++] = dp[j-b[i]][z]+a[i];
                }

                sort(c, c+w, cmp);
                w = unique(c, c+w) - c; 
                for(int t=0; t<k && t<w; t++) ///t的范围， 既不能大于 k，也不能大于 w
                    dp[j][t] = c[t];
            }
        }

        printf("%d
", dp[v][k-1]);

    }
    return 0;
}