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

4 6
1 4
2 6
3 12
2 7

Sample Output

23

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：

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int n,m,i,j;
    int a[5000],b[5000];
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0; i<n; i++)
            scanf("%d%d",&a[i],&b[i]);
        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
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
            for(j=m; j>=a[i]; j--)
                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
        printf("%d
",dp[m]);
    }
    return 0;
}