P2871 [usaco 07dec] Charm Bracelet, p2871usaco 07dec

P2871 [usaco 07dec] Charm Bracelet, p2871usaco 07dec
Description

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 'inclurability '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.

There are N items and a backpack with a capacity of V. The weight of the I-th item is c [I], and the value is w [I]. Solving which items are loaded into a backpack can make the total weight of these items not exceed the size of the backpack, and the total value is the largest.

Input/Output Format Input Format:

• 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 Format:

• Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Input and Output sample Input example #1:

4 61 42 63 122 7

Output sample #1:

23

Although it is a bare backpack, but this cannot be two-dimensional, it will burst
Then I pushed a one-dimensional one by myself ,,
It seems that it can be shorter...

 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 }