具有“至少 X 值”约束的背包

Question

您将如何解决背包的这种变体？

您有 n 个对象 (x1,...xn)，每个对象都有成本 ci 和值 vi (1<=i<=n) 以及一个附加约束 X，它是所选项目值的下限。找到 x1,...,xn 的子集，使值至少为 X 的项目的成本最小化。

我试图通过动态编程来解决这个问题，我想到的是将常用的表修改为 K[n,c,X] ，其中 X 是我需要达到的最小值，但这似乎对我无济于事。有什么好主意吗？

Answer 1

这可以像我们做背包问题一样完成，在每个索引处，我们尝试在背包内放入一个值或不放入一个值，这里背包的大小没有限制，因此我们可以在背包内放入任何元素。

那么我们只需要考虑那些满足条件的解决方案size of the knapsack >= X。

背包的状态是DP[i][j]其中i是元素的索引，j是当前背包的大小，注意我们只需要考虑那些具有 的解j >= X。

下面是 C++ 中的递归动态规划解决方案：

#include <iostream>
#include <cstring>
#define INF 1000000000


using namespace std;

int cost[1000], value[1000], n, X, dp[1000][1000];

int solve(int idx, int val){
    if(idx == n){
        //this is the base case of the recursion, i.e when
        //the value is >= X then only we consider the solution
        //else we reject the solution and pass Infinity
        if(val >= X) return 0;
        else return INF;
    }
    //this is the step where we return the solution if we have calculated it previously
    //when dp[idx][val] == -1, that means that the solution has not been calculated before
    //and we need to calculate it now
    if(dp[idx][val] != -1) return dp[idx][val];

    //this is the step where we do not pick the current element in the knapsack
    int v1 = solve(idx+1, val);

    //this is the step where we add the current element in the knapsack
    int v2 = solve(idx+1, val + value[idx]) + cost[idx];

    //here we are taking the minimum of the above two choices that we made and trying
    //to find the better one, i.e the one which is the minimum
    int ans = min(v1, v2);

    //here we are setting the answer, so that if we find this state again, then we do not calculate
    //it again rather use this solution that we calculated
    dp[idx][val] = ans;

    return dp[idx][val];
}

int main(){
    cin >> n >> X;
    for(int i = 0;i < n;i++){
        cin >> cost[i] >> value[i];
    }

    //here we are initializing our dp table to -1, i.e no state has been calculated currently
    memset(dp, -1, sizeof dp);

    int ans = solve(0, 0);

    //if the answer is Infinity then the solution is not possible
    if(ans != INF)cout << solve(0, 0) << endl;
    else cout << "IMPOSSIBLE" << endl;

    return 0;
}

ideone 上的解决方案链接：http://ideone.com/7ZCW8z