poj -Charm Bracelet 01背包

最新推荐文章于 2019-08-06 15:11:09 发布

原创最新推荐文章于 2019-08-06 15:11:09 发布 · 314 阅读

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~背包问题~

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本文介绍了一个经典的计算机科学问题——魅力手链问题。该问题要求在给定一系列装饰物的重量和吸引力评分的情况下，在不超过总重量限制的前提下，选择装饰物以获得最大的吸引力总和。文章提供了一段代码示例，展示如何通过动态规划的方法来解决这个问题。

Charm Bracelet

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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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

数组要开足够大！！

     
      
       1
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       #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int n,v[3410],t,m;
    int value[3410],rad[12900];
    while(~scanf("%d%d",&n,&m))
    {
    	
        memset(rad,0,sizeof(rad));
        for(int i=1;i<=n;i++)
        scanf("%d%d",&v[i],&value[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=v[i];j--)
            rad[j]=max(rad[j-v[i]]+value[i],rad[j]);
        }
        printf("%d\n",rad[m]);
    }
    return 0;
}