POJ 3624 Charm Bracelet(01背包裸题)

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 38909		Accepted: 16862

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

4 6
1 4
2 6
3 12
2 7

Sample Output

23

Source

USACO 2007 December Silver

题目链接：http://poj.org/problem?id=3624

分析：01背包裸题，顺带敲了一遍，做个复习吧，怕忘了！详解请参看我的博客http://www.cnblogs.com/ECJTUACM-873284962/p/6815610.html，里面有对01背包的完全介绍以及用法！

下面给出AC代码：

#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
int w[35000],d[35000],dp[35000];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d%d",&w[i],&d[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=w[i];j--)
            {
                dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}