hdu 2636 Bone Collector II 01揹包（第k優解問題）

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

百萬秦關終屬楚

 

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k優解問題，以前沒有遇到過，既然看見了就做做看

題意：有n件物品，揹包體積爲v，給出一行價值和一行花費，求第k優解，每個物品只能取一次。

思路：不考慮k優解，顯然是個簡單的01揹包，1維的數組足夠表示。即便要求k優解，在k<=30的條件下，此題再加1維也沒什麼好說的，用dp[j][k]表示揹包體積爲j時的k優解，一開始不知道怎麼轉移，原來無非是記錄當前每種可能的解，然後從大到小插進數組就行了。時間複雜度就是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;
}