《动态规划》—HDU2639 Bone CollectorII(01背包+第K有解)

    xiaoxiao2021-04-18  56

    Bone Collector II

     

    Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4662    Accepted Submission(s): 2429

     

     

    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

     

     

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    第k大 背包问题,就是开多一维去记录前k大的最优值,即p[v][k]表示装了v体积的东西获得第k大的价值网上有个比喻用得好,要计算整个年级的前n名,可以拿每班的前n名出来排序包也是这样,数组a[], b[]来保存 p[j - w[i]][1 -> k] + v[i] , 以及p[j][1 - >k] 这些值然后将两个数组里面的值按大到小的顺序重新写入 p[j][1 -k]中这样就更新了p[i]数组了 #include<cstdio> #include<cstring> using namespace std; const int maxn=1005 int dp[maxn][35],v[105],w[105],A[35],B[35]; int main(){ int n,V,t,k; scanf("%d",&t); while(t--){ scanf("%d%d%d",&n,&V,&k); for(int i=1;i<=n;i++) scanf("%d",v+i); for(int i=1;i<=n;i++) scanf("%d",w+i); memset(dp,0,sizeof(dp)); //初始化 int a,b,c,kk; for(int i=1;i<=n;i++) for(int j=V;j>=w[i];j--){ for(kk=1;kk<=k;kk++){ A[kk]=dp[j][kk]; B[kk]=dp[j-w[i]][kk]+v[i]; } A[kk]=B[kk]=-1; a=b=c=1; while(c<=k&&(A[a]!=-1||B[b]!=-1)){ if(A[a]>B[b]) dp[j][c]=A[a++]; else dp[j][c]=B[b++]; if(dp[j][c]!=dp[j][c-1]) c++; } } printf("%d\n",dp[V][k]); } return 0; }

     

     

     

     

     

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