HDU 2639 Bone Collector II

Bone Collector II

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on  HDU. Original ID: 2639
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

百万秦关终属楚

 

解题：dp背包。。。每次都排下序。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 1005;

18 int dp[maxn][75],v[maxn],c[maxn],tmp[maxn*10];

19 int N,V,K;

20 int main() {

21     int t,cnt,z;

22     scanf("%d",&t);

23     while(t--){

24         scanf("%d %d %d",&N,&V,&K);

25         for(int i = 1; i <= N; i++)

26             scanf("%d",v+i);

27         for(int i = 1; i <= N; i++)

28             scanf("%d",c+i);

29         memset(dp,0,sizeof(dp));

30         for(int i = 1; i <= N; i++){

31             for(int j = V; j >= c[i]; j--){

32                 cnt = 0;

33                 for(int k = 1; k <= K; k++){

34                     tmp[cnt++] = dp[j][k];

35                     tmp[cnt++] = dp[j-c[i]][k]+v[i];

36                 }

37                 sort(tmp,tmp+cnt);

38                 z = 1;

39                 for(int i = cnt-1; i >= 0; i--){

40                     if(z > K) break;

41                     if(i == cnt-1 || tmp[i] != tmp[i+1])

42                         dp[j][z++] = tmp[i];

43                 }

44             }

45         }

46         printf("%d\n",dp[V][K]);

47     }

48     return 0;

49 }

View Code