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：

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 }