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 ;
}

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