poj 3624 Charm Bracelet（01背包）

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29295		Accepted: 13143

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 W_i (1 ≤ W_i≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

Output

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

Sample Input

Sample Output

Source

USACO 2007 December Silver

一道最简单的01背包，特别简单，而且好理解。

题意：第一行第一个数为有几颗珠子，第二个数字表示为最大珠子和的重量，后面每一行表示一个珠子，第一个数为重量，第二数为价值，在最大重量的限制下，求最大限制。

附上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int max(int a,int b)
 6 {
 7     return a>b?a:b;
 8 }
 9 int main()
10 {
11     int n,m,i,j;
12     int a[5000],b[5000];
13     while(~scanf("%d%d",&n,&m))
14     {
15         for(i=0; i<n; i++)
16             scanf("%d%d",&a[i],&b[i]);
17         int dp[13000];   //dp数组的大小wa了一次，要注意必须和最大重量相同，而不是最大珠子个数
18         memset(dp,0,sizeof(dp));
19         for(i=0; i<n; i++)
20             for(j=m; j>=a[i]; j--)
21                 dp[j]=max(dp[j],dp[j-a[i]]+b[i]);   //比较加上这颗珠子和不加的价值谁更大，记录大的那个
22         printf("%d\n",dp[m]);
23     }
24     return 0;
25 }

poj 3624 Charm Bracelet（01背包）

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