Bone Collector II
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
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6 #define inf 0x3f3f3f3f
7 typedef long long ll;
8 const int maxn=1100;
9 int dp[maxn][maxn];
10 int val[maxn],w[maxn];
11 int casen,n,V,k;
12 int a[maxn],b[maxn];
13 int main()
14 {
15 cin>>casen;
16 while(casen--)
17 {
18 scanf("%d%d%d",&n,&V,&k);
19 for(int i=0;i<n;i++)
20 scanf("%d",&val[i]);
21 for(int i=0;i<n;i++)
22 scanf("%d",&w[i]);
23 memset(dp,0,sizeof(dp));
24 if(k==0)
25 {
26 printf("0
");
27 continue;
28 }
29 for(int i=0;i<n;i++)
30 {
31 for(int j=V;j>=w[i];j--)
32 {
33 for(int m=1;m<=k;m++)
34 {
35 a[m]=dp[j][m];
36 b[m]=dp[j-w[i]][m]+val[i];
37 }
38 int x=1,y=1,w=1;
39 a[k+1]=b[k+1]=-inf;
40 while(w<=k&&(x<=k||y<=k))
41 {
42 if(a[x]>b[y])
43 {
44 dp[j][w]=a[x++];
45 }
46 else
47 {
48 dp[j][w]=b[y++];
49 }
50 if(w==1||dp[j][w]!=dp[j][w-1])//去重
51 {
52 w++;
53 }
54 }
55 }
56 }
57 printf("%d
",dp[V][k]);
58 }
59 }
60 /*和01背包的思路一样，j从V倒着循环，越循环得到的当前体积下的最优解值越大。
61 dp[j][m]表示的是在体积为j的情况下第m大最优解,而这第m大最优解与dp[j][m]和dp[j-w[i]][m]+val[i]有关，至于
62 为什么不写成dp[j][m]=max(dp[j][m],dp[j-w[i]][m]+val[i]),而是要开a和b两个数组，是因为比如说有两组从大到小的数据：
63 第一组 第二组
64 10 8
65 5 6
66 1 2
67 max(dp[j][m],dp[j-w[i]][m]+val[i])在算m=1时，得到10便会把8掩盖，因而应该开两个数组分别记录dp[j][m]和dp[j-w[i]][m]+val[i]的m大值
68 最后以10 8 6 5 2 1的顺序赋值给dp[j][1],dp[j][2].....
69 */