HDU - 2639 -- Bone Collector II -- 01背包变形（K优解）

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 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：
　　求01背包中， 价值为第 K 大的解。

思路：
　　在背包为 j 的时候，记录下选择第 i 个物品和不选择第 i 个物品的时候的值，然后将两个值的最大值“归并”到 dp 数组中，
　　参考大神讲解K优值问题

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 1010
using namespace std;
int volume[maxn];
int value[maxn];
int chose[maxn]; //选择第 i 个物品的时候的从大到小排列
int not_chose[maxn]; //不选择第 i 个物品的时候的从大到小排列
int dp[maxn][32]; //dp[][k] 表示第k个最优解
int main()
{
    int N,K,V;
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&N,&V,&K);
        memset(dp,0,sizeof(dp));
        for(int i = 1;i <= N;i++)scanf("%d",&value[i]);
        for(int i = 1;i <= N;i++)scanf("%d",&volume[i]);
        for(int i = 1;i <= N;i++){
            for(int j = V;j >= volume[i];j--){
                int k = 0;
                for(k = 0;k <= K;k++){
                    chose[k] = dp[j-volume[i]][k]+value[i];
                    not_chose[k] = dp[j][k];
                }
                int x = 1,y = 1,z = 1;
                chose[k] = -1;
                not_chose[k] = -1;
                while(z <= K && (chose[x]!=-1 || not_chose[y]!=-1)) {
                    if(chose[x] > not_chose[y])dp[j][z] = chose[x++];
                    else dp[j][z] = not_chose[y++];
                    if(dp[j][z] != dp[j][z-1])z++;
                }
            }
        }
        printf("%d\n",dp[V][K]);
    }
    return 0;
}

 

int k = 0;
for(k = 0;k <= K;k++){
    chose[k] = dp[j-volume[i]][k]+value[i];  //记录下选择第 i 个物品的时候的价值
    not_chose[k] = dp[j][k];　　　　　　　　　  //记录下不选择第 i 个物品的时候的价值
}
int x = 1,y = 1,z = 1;
chose[k] = -1;
not_chose[k] = -1;
while(z <= K && (chose[x]!=-1 || not_chose[y]!=-1)) {   
    if(chose[x] > not_chose[y])dp[j][z] = chose[x++];  //将两者合并，归并
    else dp[j][z] = not_chose[y++];
    if(dp[j][z] != dp[j][z-1])z++;  
}

//核心代码