Bessie has gone to the mall’s jewelry store and spies a charm bracelet. Of course, she’d like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a ‘desirability’ factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
Input
- Line 1: Two space-separated integers: N and M
- Lines 2…N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
Output
- Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
Sample Input
4 6
1 4
2 6
3 12
2 7
Sample Output
23
思路:表示在前i种物品中取若干种,在其总体积不超过M的条件下所能获得的最大价值。
在这里插入代码片
#include"cstdio"
#include"cstring"
#include"algorithm"
#include"cmath"
using namespace std;
int dp[12905];
int w[5000],v[5000];
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
for(int i=1;i<=n;i++)
{
scanf("%d %d",&w[i],&v[i]);
}
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
for(int j=m;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
}
}
printf("%d\n",dp[m]);
}
return 0;
}