HDU 2639 Bone Collector II(01背包变形【第K大最优解】)

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4739 Accepted Submission(s): 2470

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

Author

teddy

Source

百万秦关终属楚

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

题目大意：

见之前的收集骨头的博客，题意类似，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

输入包括多组样例，第一行输入一个T,样例的个数，接下来每个样例都有三行，第一行包括三个整数，N,V,K,分别代表骨头的个数，背包的容量，我们需要输出的第K个最大值，第二行包括N个数，分别代表骨头的数量和接下来一行有N个数，分别表示每种骨头的价值。

输出第K个最大价值，每个样例输出一行

思路：简单的01背包基础上做，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！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int w[110];
 4 int v[110];
 5 int dp[1010][35];
 6 int d1[1010];
 7 int d2[1010];
 8 int main()
 9 {
10     int t,n,m,k,x,y,z,p;
11     scanf("%d",&t);
12     while(t--)
13     {
14         memset(w,0,sizeof(w));
15         memset(v,0,sizeof(v));
16         memset(dp,0,sizeof(dp));
17         memset(d1,0,sizeof(d1));
18         memset(d2,0,sizeof(d2));
19         scanf("%d%d%d",&n,&m,&k);
20         for(int i=1;i<=n;i++)
21         scanf("%d",&v[i]);
22         for(int i=1;i<=n;i++)
23         scanf("%d",&w[i]);
24         for(int i=1;i<=n;i++)//01背包变形
25         {
26             for(int j=m;j>=w[i];j--)
27             {
28                 for(p=1;p<=k;p++)
29                 {
30                     d1[p]=dp[j][p];
31                     d2[p]=dp[j-w[i]][p]+v[i];
32                 }
33                 d1[p]=d2[p]=-1;
34                 x=y=z=1;
35                 while((d1[x]!=-1||d2[y]!=-1)&&z<=k)
36                 {
37                     if(d1[x]>d2[y])
38                     {
39                         dp[j][z]=d1[x];
40                         x++;
41                     }
42                     else
43                     {
44                         dp[j][z]=d2[y];
45                         y++;
46                     }
47                     if(dp[j][z-1]!=dp[j][z])
48                     z++;
49                 }
50             }
51         }
52         printf("%d\n",dp[m][k]);
53     }
54     return 0;
55 }