HDU 2639 Bone Collector II【01背包 + 第K大价值】

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 2 31).

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

【题意】： 给定背包容量，骨头的个数和每个骨头的价值，这次不是求在背包容量允许的情况下，最多装的价值，而是求在背包容量内，可以装的第k大价值，如果没有第k个最大值，那么输出0。

【分析】：

求背包的第k大方案，只需在状态上加入一维dp[j][k]表示前i个物品装入容量为j的背包的第k大的方案，用两个数组辅助保存下装和不装两种选择下的前k大方案，再最后合并起来得到最终结果 。

要求的是第K个最大值，那么不用 dp[j]=max(dp[j],dp[j-w[i]]+v[i])的状态转移方程，而是将两个值都记录下来，用for循环走一遍，记录下，容量为1到M的各个最大价值，dp[i][j]表示当背包容量为i时的第j个最大价值，最后只需要输出dp[m][k]即可

【精彩讲解】：HDU2639 01背包 第K优决策

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn =  1e5 + 5;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
/*
总的复杂度是O(VNK)
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
*/
int n,m,K;
#define S 100000
#define M 200000
int w[1005],v[1005];
int a[1005],b[1005];
int dp[1005][1005];//设dp[j][k]为容量为j的背包所获得的第k大价值
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));

        cin>>n>>m>>K;
        for(int i=1;i<=n;i++) cin>>v[i];
        for(int i=1;i<=n;i++) cin>>w[i];
        for(int i=1;i<=n;i++){
            for(int j=m;j>=w[i];j--){
                for(int k=1;k<=K;k++){ //用两个数组辅助保存下装和不装两种选择下的前k大方案
                    a[k] = dp[j][k];
                    b[k] = dp[j-w[i]][k] + v[i];
                }
                //a[K+1]=b[K+1]=-1;
                int x=1,y=1,z=1;
                while(z<=K && (x<=K||y<=K)){ //二路归并
                    if(a[x]>=b[y]) //不能去掉等于
                        dp[j][z]=a[x++];
                    else
                        dp[j][z]=b[y++];
                    if(dp[j][z-1] != dp[j][z]) z++; //去重
                }
            }
        }
        cout<<dp[m][K]<<endl;
    }
    return 0;
}