[DP] POJ - 3624 Charm Bracelet

最新推荐文章于 2019-08-25 16:44:00 发布

原创最新推荐文章于 2019-08-25 16:44:00 发布 · 215 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#poj #dp

算法_动态规划专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一个经典的01背包问题，给出了完整的代码实现，并详细解释了如何在给定的重量限制下选择一系列具有不同权重和价值的物品，以达到最大的总价值。此问题常见于计算机科学中的动态规划领域。

Charm Bracelet
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 41212 Accepted: 17929

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 Wi (1 ≤ Wi ≤ 400), a ‘desirability’ factor Di (1 ≤ Di ≤ 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: Wi and Di

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

题意: 01背包问题,
输入的第一行有两个数，第一个是代表有n件物品，第二个表示背包的重量限制。接下来有n行数据，每行数据包含两个数字。第一个表示这件物品的重量，第二个数字表示这件物品的价值。
我们要做的是求出在重量限制下，背包最多能装多少价值的物品。每件物品只有一件。

//已AC代码
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <complex>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cassert>
using namespace std;
#define N 1050017
int main()
{
    int dp[30000];
    int n,m;
    int w[4000];
    int d[4000];
    int i, j;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        for (i=1;i<=n;i++)
            cin>>w[i]>>d[i];
        memset(dp,0,sizeof(dp));
        for (i=1;i<=n;i++)
        {
            for (j=m;j>=w[i];j--)
            {
                dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
            }
        }
        cout<<dp[m]<<endl;
    }
}