hdu 2636 Bone Collector II 01揹包（第k优解问题）

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2286    Accepted Submission(s): 1197

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

 

Recommend

teddy   |   We have carefully selected several similar problems for you:  2602 2159 1203 2955 1114 

k优解问题，以前没有遇到过，既然看见了就做做看

题意：有n件物品，揹包体积为v，给出一行价值和一行花费，求第k优解，每个物品只能取一次。

思路：不考虑k优解，显然是个简单的01揹包，1维的数组足够表示。即便要求k优解，在k<=30的条件下，此题再加1维也没什么好说的，用dp[j][k]表示揹包体积为j时的k优解，一开始不知道怎么转移，原来无非是记录当前每种可能的解，然后从大到小插进数组就行了。时间复杂度就是o(nvk)，最大也就3*10^6。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
const int N=105;
int t,v,n,k;
int p[N],w[N],dp[1005][35],a[35],b[35];
int main()
{
    scanf("%d",&t);
    while(t--){
        memset(dp,0,sizeof(dp));
        scanf("%d%d%d",&n,&v,&k);
        for(int i=1;i<=n;i++)scanf("%d",&p[i]);
        for(int i=1;i<=n;i++)scanf("%d",&w[i]);
        for(int i=1;i<=n;i++){
            for(int j=v;j>=w[i];j--){
               for(int m=1;m<=k;m++){
                    a[m]=dp[j][m];
                    b[m]=dp[j-w[i]][m]+p[i];
                }
                int x=1,y=1,w=1;
                a[k+1]=b[k+1]=-1;
                while(w<=k&&(x<=k||y<=k)){
                    if(a[x]>b[y])dp[j][w]=a[x++];
                    else dp[j][w]=b[y++];
                    if(w==1||dp[j][w]!=dp[j][w-1])w++;
                }
            }
        }
        printf("%d\n",dp[v][k]);
    }
    return 0;
}