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HDU 2639 Bone Collector II

2024-07-10 16:51:05   222人阅读

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
百万秦关终属楚
 

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解题思路：求背包的第k大方案，只需在状态上加入一维dp[j][k]表示前i个物品装入容量为j的背包的第k大的方案，用两个数组辅助保存下装和不装两种选择下的前k大方案，再最后合并起来得到最终结果
#include <iostream>
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<n;i++)
			scanf("%d",val+i);
		for(i=0;i<n;i++)
			scanf("%d",cost+i);
		for(i=0;i<n;i++)
			for(j=v;j>=cost[i];j--){
				for(p=0;p<k;p++){
					a[p]=dp[j][p];
					b[p]=dp[j-cost[i]][p]+val[i];
				}
				int s=0,t=0;
				p=0;
				a[k]=b[k]=-1;
				while(p<k&&!(a[s]==-1&&b[t]==-1)){
					if(a[s]>b[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;
}


 


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