0-1 Knapsack Problem(01背包模板)

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本文详细解析了01背包问题的算法实现，通过一个具体的示例介绍了如何在不超过背包容量的条件下，选择物品以达到价值最大化。文章提供了一个C++实现的代码模板，帮助读者理解和应用01背包算法。

滴答滴答---题目链接

You have N items that you want to put them into a knapsack. Item ihas value vi and weight wi.

You want to find a subset of items to put such that:

The total value of the items is as large as possible.
The items have combined weight at most W, that is capacity of the knapsack.

Find the maximum total value of items in the knapsack.

Input

N W
v1 w1
v2 w2
:
vN wN

The first line consists of the integers N and W. In the following lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Constraints

1 ≤ N ≤ 100
1 ≤ vi ≤ 1000
1 ≤ wi ≤ 1000
1 ≤ W ≤ 10000

Sample Input 1

Sample Output 1

Sample Input 2

2 20
5 9
4 10

Sample Output 2

n,m

a,w;

题意：n个产品，不超过m重量，接下来a代表一个产品的价格，w表示一个产品的重量，输出在不超过m重量的前提下，最大得到的价格

01背包模板

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int inf=0x3f3f3f;
int dp[10010];
int a[10100],w[10010];
int main()
{memset(dp,0,sizeof(dp));
    int m,n;
    cin>>n>>m;
    for(int i=1; i<=n; i++)
        cin>>a[i]>>w[i];
    for(int i=1; i<=n; i++)
        for(int j=m; j>=w[i]; j--)
            dp[j]=max(dp[j],dp[j-w[i]]+a[i]);
    printf("%d\n",dp[m]);
    return 0;
}