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  • poj 3624 Charm Bracelet

    http://poj.org/problem?id=3624

    一维数组的01背包,v一定要逆序枚举,不然会多次放同一种物品。

                                                                                      Charm Bracelet
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 24456   Accepted: 11031

    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 '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.

    Input

    * 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

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

    Sample Input

    4 6
    1 4
    2 6
    3 12
    2 7

    Sample Output

    23
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    int c,n;
    int v[3412],w[3412],f[12888];
    void backpack()
    {
        memset(f,0,sizeof(f));
        int i,j;
        for(i=1;i<=n;i++)
          for(j=c;j>=0;j--)
             if(j>=v[i])
                f[j]=max(f[j],f[j-v[i]]+w[i]);
         printf("%d
    ",f[c]);
    
    }
    int main()
    {
        int i;
        scanf("%d%d",&n,&c);
        for(i=1;i<=n;i++)
          scanf("%d%d",&v[i],&w[i]);
        backpack();
        return 0;
    
    }
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  • 原文地址:https://www.cnblogs.com/cancangood/p/4165930.html
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