HDU2639Bone Collector II[01背包第k优值]

Posted 2020-08-09 Candy?

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4229    Accepted Submission(s): 2205

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

---

每个状态保存1到k优值，转移时给f[j][]和f[j-v]+w二路归并一下就好了

复杂度O(nvk)

 

//
//  main.cpp
//  hdu2639
//
//  Created by Candy on 9/22/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=105,V=1005,K=35,INF=1e9+5;
int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();}
    return x*f;
}
int T,n,m,k,w[N],v[N];
int f[V][K],a[K],b[K];
void dp(){
    memset(f,0,sizeof(f));
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++)
        for(int j=m;j>=v[i];j--){
            for(int z=1;z<=k;z++) {a[z]=f[j][z];b[z]=f[j-v[i]][z]+w[i];}
            int x=1,y=1,z=1;
            while(z<=k&&(x<=k||y<=k)){
                if(a[x]>=b[y]) f[j][z]=a[x++];
                else f[j][z]=b[y++];
                if(f[j][z]!=f[j][z-1]) z++;
            }
        }
}
int main(int argc, const char * argv[]) {
    T=read();
    while(T--){
        n=read();m=read();k=read();
        for(int i=1;i<=n;i++) w[i]=read();
        for(int i=1;i<=n;i++) v[i]=read();
        dp();
        printf("%d\n",f[m][k]);
    }
    
    return 0;
}