HDU 2639 Bone Collector II

Bone Collector IITime Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 2089 Accepted Submission (s): 1097


Problem DescriptionThe 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 = 1, 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 instances.
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 231 ).

Sample Input

35 10 21 2 3 4 55 4 3 2 15 10 121 2 3 4 55 4 3 2 15 10 161 2 3 4 55 4 3 2 1

Sample Output

1220

Authorteddy
Source million Qin Guan Chu
Recommendteddy | We have carefully selected several similar problems for you: 2602 2159 1203 2955 Solution: Find the k solution of the backpack, you only need to add one-dimensional dp [j] [k] to the status, which indicates the k-large solution for loading the first I item into a backpack with a capacity of j, use two arrays to save the first k large scheme of the two options, and then combine them to get the final result.

#include 
 
  using namespace std;bool cmp(int a,int b){return a>b;}int main(){int i,j,t,n,v,k,p;int cost[105],val[105],dp[1005][35],a[35],b[35];//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);scanf("%d",&t);while(t--){memset(dp,0,sizeof(dp));scanf("%d%d%d",&n,&v,&k);for(i=0;i
  
   =cost[i];j--){for(p=0;p
   
    b[t])dp[j][p]=a[s++];elsedp[j][p]=b[t++];if(dp[j][p]!=dp[j][p-1])p++;}}printf("%d\n",dp[v][k-1]);}return 0;}