hdu2639 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.

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.

One integer per line representing the K-th maximum of the total value (this number will be less than 2³¹).

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

12
2
0
这题很巧妙，用的是01背包思想，普通的01背包求的是最优解，但是题目要求的是第K最大值，所以我们要多加一维状态，即dp[j][k]表示质量为j的第k大值,那么用a[],b[]记录dp[j][h]和dp[j-w[i]][h]+v[i]（h从1~k）。然后通过这两个数组求出k个依次排列的最大值dp[j][k].
#include<stdio.h>
#include<string.h>
int v[106],w[105],dp[1006][50];
int main()
{
	int n,m,i,j,T,a[200],b[200],h,k,t,x,y;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d",&n,&m,&k);
		for(i=1;i<=n;i++){
			scanf("%d",&v[i]);
		}
		for(i=1;i<=n;i++){
			scanf("%d",&w[i]);
		}
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++){
			for(j=m;j>=w[i];j--){
				for(h=1;h<=k;h++){
					a[h]=dp[j][h];
					b[h]=dp[j-w[i]][h]+v[i];
				}
				a[k+1]=-1;b[k+1]=-1;
				h=0;x=y=1;
				while(h<k && (x<=k || y<=k)){
					if(a[x]>b[y]){
						t=a[x++];
					}
					else t=b[y++];
					if(h==0 ||(h>0 && dp[j][h]!=t)){
						h++;dp[j][h]=t;
					} 
				}
			}
		}
		printf("%d\n",dp[m][k]);
	}
	return 0;
}