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hdu2639

2024-07-04 22:03:06   224人阅读

这是一个典型的求第k优解的问题。。背包九讲上面都有。。。所有我只讲一下我的思路。。求第k优解就是在最优解的复杂度上增加了一个k，每次用一个物品去更新dp[v]的时候用一个数组保存才来。。然后排序，题目说要去重，那就很好办了。。然后经过N个物品的更新，那么最后的答案就是打dp[V][K]..

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2639

题目为：

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2037    Accepted Submission(s): 1058


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

代码为：

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[1005][40],cost[105],value[105],ans[10000];
int N,V,K;
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int t,i,j,cnt,pos,k;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d%d%d",&N,&V,&K);
         for(i=1;i<=N;i++)
            scanf("%d",&value[i]);
         for(i=1;i<=N;i++)
            scanf("%d",&cost[i]);
         for(i=1;i<=N;i++)
             for(j=V;j>=cost[i];j--)
                    {
                        cnt=1;
                        for(k=1;k<=K;k++)
                       {
                         ans[cnt++]=dp[j][k];
                         ans[cnt++]=dp[j-cost[i]][k]+value[i];
                       }
                       sort(ans+1,ans+cnt,cmp);
                       int x=1;
                       for(k=1;k<cnt-1;k++)
                       {
                           if(x>K)
                            break;
                           if(ans[k]!=ans[k+1])
                            dp[j][x++]=ans[k];
                       }
                    }
            printf("%d\n",dp[V][K]);
    }
    return 0;
}




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