Hdu 2639 Bone Collector II

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本文探讨了骨收集器II问题，即在给定的骨头数量、背包容量和第K大需求下，如何计算最大的骨头总价值。通过分析案例输入输出，学习如何实现01背包算法来解决此问题。

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Bone Collector II

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 921 Accepted Submission(s): 440

题目链接

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

Sample Output

Author

teddy

Source

百万秦关终属楚

Recommend

teddy

求01背包的第k大解。

#include<stdio.h>
#include<string.h>

int w[101],v[101];
int a[31],b[31];
int dp[1001][31];

int main(void)
{
	int T,N,V,K,i,j,m,n_a,n_b,n_c;
//	freopen("d:\\in.txt","r",stdin);
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d",&N,&V,&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=V;j>=w[i];j--)
			{
				for(m=1;m<=K;m++)
					a[m]=dp[j-w[i]][m]+v[i];
				for(m=1;m<=K;m++)
					b[m]=dp[j][m];
				a[m]=b[m]=-1;
				n_a=n_b=n_c=1;
				while(n_c<=K && (n_a<= K || n_b<=K))
				{
					if(a[n_a]>=b[n_b])
						dp[j][n_c]=a[n_a++];
					else
						dp[j][n_c]=b[n_b++];
					if(dp[j][n_c]!=dp[j][n_c-1])
						n_c++;
				}
			}
		printf("%d\n",dp[V][K]);
	}
	return 0;
}