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.
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.
InputThe 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.
OutputOne 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
题目意思:0/1背包的变形,求第N优解
解题思路:储存每一次的变化;
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include<math.h>
5 #include <algorithm>
6 using namespace std;
7
8 const int MAX = 2000;
9
10 int main()
11 {
12 int N;
13 cin>>N;
14 while(N--)
15 {
16 int n,m,ti;
17 scanf("%d %d %d",&n,&m,&ti);
18
19 int w[MAX],v[MAX];
20 for(int i =1;i<=n;i++)
21 scanf("%d",&v[i]);
22
23 for(int j = 1;j <=n;j++)
24 scanf("%d",&w[j]);
25
26 int tab[MAX][MAX],a[MAX],b[MAX];
27 memset(tab,0,sizeof(tab));
28 for(int i =1;i <=n;i++)
29 {
30 for(int j =m;j>=w[i];j--)
31 {
32 for(int k =1; k<=ti;k++)
33 {
34 a[k] = tab[j-w[i]][k]+v[i];
35 b[k] = tab[j][k];
36 }
37 a[ti+1]=-1;
38 b[ti+1] =-1;
39 int x,y,z;
40 x= y =z= 1;
41 while(z<=ti&&(a[x]!=-1||b[y]!=-1))
42 {
43 if(a[x]>b[y])
44 tab[j][z] = a[x++];
45 else
46 tab[j][z] = b[y++];
47
48 if(tab[j][z]!=tab[j][z-1])
49 z++;
50 }
51 }
52 }
53 cout<<tab[m][ti]<<endl;
54
55 }
56
57
58 return 0;
59 }