Bone Collector II
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3355 Accepted Submission(s):
1726
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
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还是01背包问题，不过我是第一次写这种类型的题，参考了别人的代码和思想，差不多是一道模板第k大价值题，值得一做。
更新的时候有两个数组A、B，A记录选择第i个物品的从大到小排列，B记录不选择的从大到小排列，然后合并AB，选出AB里面前k个最大的。合并到dp中。。。
题意：第一行三个数，第一个数为物体数量，第二个数总体积，第三个数第k优解。注意！接下来第二行是物体的价值，第三行是物体的体积。
求第k大优解的值是多少。
附上代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5 int main()
6 {
7 int T,i,j,n,m,v,k,kk;
8 int dp[1005][35],val[1005],vol[1005],A[35],B[35];
9 scanf("%d",&T);
10 while(T--)
11 {
12 scanf("%d%d%d",&n,&v,&k);
13 for(i=0; i<n; i++)
14 scanf("%d",&val[i]); //价值
15 for(i=0; i<n; i++)
16 scanf("%d",&vol[i]); //体积
17 memset(dp,0,sizeof(dp));
18 int a,b,c;
19 for(i=0; i<n; i++)
20 for(j=v; j>=vol[i]; j--)
21 {
22 for(kk=1; kk<=k; kk++) //从最优的向后循环
23 {
24 A[kk]=dp[j-vol[i]][kk]+val[i]; //选中i物体
25 B[kk]=dp[j][kk]; //不选i物体
26 }
27 A[kk]=-1; //-1标记为结尾
28 B[kk]=-1;
29 a=b=c=1;
30 while(c<=k&&(A[a]!=-1||B[b]!=-1))
31 {
32 if(A[a]>B[b]) //大的数排在前面
33 dp[j][c]=A[a++];
34 else
35 dp[j][c]=B[b++];
36 if(dp[j][c]!=dp[j][c-1]) //如果两数相同，则只记录一次
37 c++;
38 }
39 }
40 printf("%d\n",dp[v][k]);
41 }
42 return 0;
43 }