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 虽然是裸的背包，但是这个不能用二维，会爆
然后自己手推了一下一维的，，
貌似还能更短。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
void read(int & n)
{
char c='+';int x=;int flag=;
while(c<''||c>'')
{
c=getchar();
if(c=='-')
flag=;
}
while(c>=''&&c<='')
x=x*+(c-),c=getchar();
flag==?n=-x:n=x;
}
const int MAXN=;
int n,maxt;
struct node
{
int w;
int v;
}a[MAXN];
int dp[MAXN];
int main()
{
read(n);read(maxt);
for(int i=;i<=n;i++)
{
read(a[i].w);read(a[i].v);
}
for(int i=;i<=n;i++)
{
for(int j=maxt;j>=;j--)
{
if(a[i].w<=j)
dp[j]=max(dp[j],dp[j-a[i].w]+a[i].v);
else
dp[j]=dp[j];
}
}
cout<<dp[maxt];
return ;
}