Knapsack problem FZU - 2214 （01背包变形）_java问题which is the maximum allowed weight of a kna-优快云博客

本文介绍了一个背包问题的经典求解方法，通过动态规划算法确定在给定容量限制下所能获得的最大价值。文章提供了一段C++代码实现，展示了如何初始化状态数组，并通过迭代更新来寻找最优解。

Knapsack problem

Given a set of n items, each with a weight w[i] and a value v[i], determine a way to choose the items into a knapsack so that the total weight is less than or equal to a given limit B and the total value is as large as possible. Find the maximum total value. (Note that each item can be only chosen once).

Input

The first line contains the integer T indicating to the number of test cases.

For each test case, the first line contains the integers n and B.

Following n lines provide the information of each item.

The i-th line contains the weight w[i] and the value v[i] of the i-th item respectively.

1 <= number of test cases <= 100

1 <= n <= 500

1 <= B, w[i] <= 1000000000

1 <= v[1]+v[2]+...+v[n] <= 5000

All the inputs are integers.

Output

For each test case, output the maximum value.

Sample Input

Sample Output

我们用dp针对不同的价值计算最小的重量

当时想到了这样转变就是不知道怎么初始化了结果就没做出来哎

初始化为INF表示不存在而且由于前0个物品中什么都挑选不了，所以初始化dp[0] = 0；

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
int v[1010],w[1010];
int dp[5050];//dp[i]表示价值为i时的最小重量
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,W;
        scanf("%d%d",&n,&W);
        int V = 0;
        for(int i = 0; i < n; i++){
            scanf("%d%d",&w[i],&v[i]);
            V += v[i];
        }
        //初始化（很重要！）
        memset(dp,INF,sizeof(dp));
        dp[0] = 0;
        for(int i = 0; i < n; i++){
            for(int j = V; j >= v[i]; j--){
                dp[j] = min(dp[j],dp[j-v[i]]+w[i]);
            }
        }
        for(int i = V; i >= 0; i--){
            if(dp[i] <= W){
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}