01背包第K最优解 HDU - 2639 

╰+攻爆jí腚メ 2022-06-17 07:56 124阅读 0赞

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 2 31).  
  
  
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

#include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    int t,n,v,K;
    int dp[1005][35];
    int a[35];
    int b[35];
    int value[105];
    int weight[105];
    int main()
    {
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d%d%d",&n,&v,&K);
            for(int i=1; i<=n; i++)
                scanf("%d",&value[i]);
            for(int i=1; i<=n; i++)
                scanf("%d",&weight[i]);
    
            memset(dp,0,sizeof(dp));
            for(int i=1; i<=n; i++)
                for(int j=v; j>=weight[i]; j--) //在背包为 j 的时候，记录下选择第 i 个物品和不选择第 i 个物品的时候的值

//然后将两个值的最大值“归并”到 dp 数组中
                {
                     for(int k=1; k<=K; k++)     //多加一维用来找第K最优解
                    {
                        a[k]=dp[j-weight[i]][k]+value[i];
                        b[k]=dp[j][k];
                    }
                    a[K+1]=-1;b[K+1]=-1;
                    int x=1,y=1,c=1;
                    while((a[x]!=-1||b[y]!=-1) && c<=K)  //每一次放进一个骨头找最优解
                    {
                        if(a[x]>b[y]) dp[j][c]=a[x],x++;
                        else dp[j][c]=b[y],y++;
                        if(dp[j][c]!=dp[j][c-1]) c++;   //找到最优解才更新
                    }
    
                }
            printf("%d\n",dp[v][K]);
    
        }
    
        return 0;
    }