（01背包第k优解） Bone Collector II（hdu 2639）-白红宇

（01背包第k优解） Bone Collector II（hdu 2639）

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发布时间：2019-06-28

本文共 2173 字，大约阅读时间需要 7 分钟。

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:

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

思路：

对于求最优解的情况，我们对每一种状态只保存了该状态下的最优解，忽略了其他解，进而实现状态之间的转移，而对于求第K优解的情况呢？其实只需要保存每一种状态下的前K优解，从这K个状态进行状态间的转移，同时去重，保存当前状态的K优解即可。（感觉时间复杂度还是挺高的）

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             using namespace std;const int N = 6600;const int INF = 0x3fffffff;const long long MOD = 1000000007;typedef long long LL;#define met(a,b) (memset(a,b,sizeof(a)))int dp[N][35];int a[N], b[N], c[N];///dp[j][k] 代表容量为 j 的背包的第 k+1 优解int cmp(int a, int b){ return a > b;}int main(){ int T; scanf("%d", &T); while(T--) { int i, j, k, n, v; scanf("%d%d%d", &n, &v, &k); met(a, 0); met(b, 0); met(dp, 0); for(i=1; i<=n; i++) scanf("%d", &a[i]); for(i=1; i<=n; i++) scanf("%d", &b[i]); for(i=1; i<=n; i++) { for(j=v; j>=b[i]; j--) { int w = 0; for(int z=0; z

转载于:https://www.cnblogs.com/YY56/p/5476797.html

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