Bone Collector II

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


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 百萬秦關終屬楚
Recommend teddy | We have carefully selected several similar problems for you: 2602 2159 1203 2955 1114 解題思路：求背包的第k大方案，只需在狀態上加入一維dp[j][k]表示前i個物品裝入容量為j的背包的第k大的方案，用兩個數組輔助保存下裝和不裝兩種選擇下的前k大方案，再最後合並起來得到最終結果

#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;pb[t])
						dp[j][p]=a[s++];
					else
						dp[j][p]=b[t++];
					if(dp[j][p]!=dp[j][p-1])
						p++;
				}
			}
			printf("%d\n",dp[v][k-1]);
	}
	return 0;
}