P2871 [USACO07DEC]手链Charm Bracelet
P2871 [USACO07DEC]手链Charm Bracelet
attack
发布于 2018-04-13 11:24:04
发布于 2018-04-13 11:24:04
文章被收录于专栏：数据结构与算法
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 }本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。
原始发表：2017-06-22 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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