题目描述
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.
有N件物品和一个容量为V的背包。第i件物品的重量是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量，且价值总和最大。
输入输出格式
输入格式：
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
输出格式：
- Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
输入输出样例
输入样例#1：
4 6
1 4
2 6
3 12
2 7
输出样例#1：
23 思路：
裸01背包； 来，上代码：
#include <cstdio>
#include <iostream>
#include <algorithm> using namespace std; int if_z,dp[],n,m,vi,ci; char Cget; inline void in(int &now)
{
now=,if_z=,Cget=getchar();
while(Cget>''||Cget<'')
{
if(Cget=='-') if_z=-;
Cget=getchar();
}
while(Cget>=''&&Cget<='')
{
now=now*+Cget-'';
Cget=getchar();
}
now*=if_z;
} int main()
{
in(n),in(m);
while(n--)
{
in(vi),in(ci);
for(int i=m;i>=vi;i--) dp[i]=max(dp[i],dp[i-vi]+ci);
}
cout<<dp[m];
return ;
}