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

• 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

• 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

4 61 42 63 122 7

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。。。

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 void read(int & n) 7 { 8     char c=‘+‘;int x=0;int flag=0; 9     while(c<‘0‘||c>‘9‘)10     {11         c=getchar();12         if(c==‘-‘)13         flag=1;14     }15     while(c>=‘0‘&&c<=‘9‘)16     x=x*10+(c-48),c=getchar();17     flag==1?n=-x:n=x;18 }19 const int MAXN=1000001;20 int n,maxt;21 struct node22 {23     int w;24     int v;25 }a[MAXN];26 int dp[MAXN];27 int main()28 {29     read(n);read(maxt);30     for(int i=1;i<=n;i++)31     {32         read(a[i].w);read(a[i].v);33     }34     for(int i=1;i<=n;i++)35     {36         for(int j=maxt;j>=0;j--)37         {38             if(a[i].w<=j)39             dp[j]=max(dp[j],dp[j-a[i].w]+a[i].v);40             else41             dp[j]=dp[j];42         }43     }44     cout<<dp[maxt];45     return 0;46 }