Bone Collector II（01背包第K优解）

最新推荐文章于 2019-10-25 00:24:50 发布

LBJHan

最新推荐文章于 2019-10-25 00:24:50 发布

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分类专栏： dp(动态规划)

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本文介绍了一种解决第K优解问题的方法，通过在动态规划基础上增加一维数组来存储不同价值的解，并通过样例说明如何实现从最优解到第K优解的过程。

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Bone Collector II

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5167 Accepted Submission(s): 2726

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 2 ³¹).

Sample Input

Sample Output

第一次接触第K优解问题，在网上搜了搜，有点明白了；

首先了解一下什么是第K优解；当我们计算最优解时我们只取我们所需要的解，这就导致我们摒弃了许多不优解，而计算第K优解时，就需要把原先摒弃的解按顺序放入一个数组；也就是说在原先求最优解的基础上再加一维数组，若当时是dp[i][j]，现在改为dp[i][j][k]；若当初为dp[i]，现在改为dp[i][k]；

现在看看代码：

#include <iostream>
#include <algorithm>
#include <string.h>

using namespace std;

int dp[1005][35];
int main(){

    int T;
    cin >> T;
    while(T--){
        int N, V, K;
        cin >> N >> V >> K;
        int val[105], v[105];
        for(int i=1; i<=N; i++)
            cin >> val[i];
        for(int i=1; i<=N; i++)
            cin >> v[i];
        memset(dp, 0, sizeof(dp));
        int a[35], b[35];
        for(int i=1; i<=N; i++){
            for(int j=V; j>=v[i]; j--){
                for(int k=1; k<=K; k++){         //所有解都存起来；
                    a[k]=dp[j][k];
                    b[k]=dp[j-v[i]][k]+val[i];
                }
                a[K+1]=-1;
                b[K+1]=-1;
                int k=1;
                int x, y;
                x=y=1;
                while(k!=K+1 && (a[x]!=-1 || b[y]!=-1)){//所有解由最优到不优排列；
                    if(a[x]>b[y]){
                        dp[j][k]=a[x];
                        x++;
                    }
                    else{
                        dp[j][k]=b[y];
                        y++;
                    }
                    if(dp[j][k]!=dp[j][k-1]) k++;  //筛选重复解；
                }
            }
        }
        cout << dp[V][K] << endl;
    }

    return 0;
}