HDU 2639 背包第k优解

Posted 2020-09-21 LuZhiyuan

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4824    Accepted Submission(s): 2514

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

 题意：

有n件物品，每件物品有价值和体积，有容量为m的背包，求能够得到的第k大的价值

代码:

//以前的dp[V]数组再加一维dp[V]K]表示V状态时第k大的值,当枚举到第i个物品时
//dp[i][V]=max(dp[i-1][V],dp[i-1][V-v])，当前状态由两个状态转移来的所以前k大的值
//也是由两个状态的前k大的值转移来的。注意本体价值重复的算一个。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=109;
const int MAXV=1009;
const int MAXK=39;
int dp[MAXV][MAXK];
int val[MAXN],vol[MAXN];
int N,V,K;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&N,&V,&K);
        for(int i=1;i<=N;i++) 
            scanf("%d",&val[i]);
        for(int i=1;i<=N;i++)
            scanf("%d",&vol[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=N;i++){
            for(int j=V;j>=vol[i];j--){
                int a1=1,a2=1,p1[39],p2[39];
                for(int c=1;c<=K;c++){
                    p1[c]=dp[j][c];
                    p2[c]=dp[j-vol[i]][c]+val[i];
                }
                p1[K+1]=p2[K+1]=-1;
                int c=0;
                while(c!=K){
                    int tmp=max(p1[a1],p2[a2]);
                    if(tmp==p1[a1]){
                        a1++;
                        if(tmp!=dp[j][c]) dp[j][++c]=tmp;
                        else if(tmp==0) dp[j][++c]=0;
                    }
                    else if(tmp==p2[a2]){
                        a2++;
                        if(tmp!=dp[j][c]) dp[j][++c]=tmp;
                        else if(tmp==0) dp[j][++c]=0;
                    }
                }
                //cout<<i<<" "<<j<<endl;
                //for(int k=1;k<=K;k++) cout<<dp[j][k]<<" ";
                //cout<<endl;
            }
        }
        printf("%d\n",dp[V][K]);
    }
    return 0;
}