POJ-3426-0-1背包Charm Bracelet-优快云博客

本博客介绍了一个解决魅力手链填充问题的方法，通过运用0-1背包算法，最大化魅力值总和，同时考虑重量限制。详细解释了输入输出格式、解题思路及代码实现。

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Charm Bracelet

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

套用0-1背包的模版即可,这是我写的最快的一次,用了20分钟,并且一次就过了.

dp[i][v]=max{dp[i-1][v],dp[i-1][v-c[i]]+w[i]}

#include<stdio.h>
#include<string.h>
int main()
{
    int i,N,M,w[3500],D[3500],dp[13000],j;
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(i=1; i<=N; i++)
        {
            scanf("%d%d",&w[i],&D[i]);
        }
        for(i=1; i<=N; i++)
        {
            for(j=M; j>=0; j--)
            {
                if(j-w[i]>=0)
                    dp[j]=dp[j]>dp[j-w[i]]+D[i]? dp[j]:dp[j-w[i]]+D[i];
                else
                    dp[j]=dp[j];
            }
        }
        printf("%d\n",dp[M]);
    }

}