FZU2214 C - Knapsack problem

最新推荐文章于 2018-09-07 09:50:58 发布

OJBFOWE

最新推荐文章于 2018-09-07 09:50:58 发布

阅读量293

点赞数

CC 4.0 BY-SA版权

分类专栏： Algorithm------动态规划--背包文章标签：动态规划

本文链接：https://blog.youkuaiyun.com/m0_37109329/article/details/78586436

Algorithm------动态规划--背包专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一种解决经典背包问题的高效算法。该算法通过动态规划寻找在不超过背包容量限制的情况下，价值总和最大的物品组合。输入包括测试案例数量、物品数量及背包容量等参数，输出则是最大可能的价值总和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

B太大，以价值为背包，求最大价值下最小体积，最后价值从大到小遍历dp，如果一旦有容积小于B,则记录价值，break；

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
int dp[5500];
int v[550],w[550];
const int inf=0x3f3f3f3f;
int main()
{
int t,n,m;
    cin>>t;
    while (t--)
    {
     int sum=0,ans=0;
      scanf("%d%d",&n,&m);
      for (int i=1;i<=n;i++)
      {
         scanf("%d%d",&w[i],&v[i]);
         sum+=v[i];
        }
         memset(dp,inf,sizeof(dp));
         dp[0]=0;
       for (int i=1;i<=n;i++)
        for (int j=sum;j>=v[i];j--)
          dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
        for (int i=sum;i>=0;i--)
        {
           if (dp[i]<=m)
           {
             ans=i;
             break;
           }
        }
        cout<<ans<<endl;
    }
    return 0;
}