• P2871 [USACO07DEC]手链Charm Bracelet


    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。。。

     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 node
    22 {
    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             else
    41             dp[j]=dp[j];
    42         }
    43     }
    44     cout<<dp[maxt];
    45     return 0;
    46 }
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/7067033.html
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