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
1
5 15
12 4
2 2
1 1
4 10
1 2
Sample Output
15
前阵子刚学的背包正好用上了,hhh,正好来记录一下,巩固巩固
思路:一看给的w[i]到了1e9,那么之前01背包的做法枚举空间就肯定不行了,那么枚举价值就完事了,让价值大的情况下占的空间尽量小。需要注意的是初始化要初始成无穷大,但是dp[0]要等于0。
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
ll dp[5005];
int w[505];
int v[505];
int main()
{
int t,n,m,sum;
scanf("%d",&t);
while(t--)
{
sum=0;
memset(dp,inf,sizeof(dp));
dp[0]=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d%d",&w[i],&v[i]);
sum+=v[i];
}
//cout<<sum<<endl;
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)
{
cout<<i<<endl;
break;
}
}
return 0;
}