POJ 3624 Charm Bracelet

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本文介绍了一种使用一维动态规划（DP）方法解决背包问题的经典案例。通过对题目要求的解析，展示了如何在给定重量限制下选择一系列具有特定权重和价值的装饰品以达到最高的总价值。同时对比了一维DP与二维DP在内存使用上的区别。

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 31350		Accepted: 13943

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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightW_i (1 ≤ W_i ≤ 400), a 'desirability' factorD_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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

Source

简单O1背包水题，温习一下，瞬秒。

此题智能用一维DP数组，二维妥妥超内存。

一维AC

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

const int MAXN=3405;
const int MAXM=12885;
int dp[MAXM];
int weight[MAXN];
int value[MAXN];

int main()
{
        int n,m;
        while(scanf("%d%d",&n,&m)>0)
        {
                memset(dp,0,sizeof(dp));
                for(int i=1;i<=n;i++)
                        scanf("%d%d",&weight[i],&value[i]);
                for(int i=1;i<=n;i++)
                        for(int j=m;j>=weight[i];j--)
                                dp[j]=max(dp[j-weight[i]]+value[i],dp[j]);
                printf("%d\n",dp[m]);
        }
        return 0;
}

二维数组，方法正确，但此题会超内存

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

const int MAXN=3405;
const int MAXM=12885;
int dp[MAXN][MAXM];
int weight[MAXN];
int value[MAXN];

int main()
{
        int n,m;
        while(scanf("%d%d",&n,&m)>0)
        {
                memset(dp,0,sizeof(dp));
                for(int i=1;i<=n;i++)
                        scanf("%d%d",&weight[i],&value[i]);
                for(int i=1;i<=n;i++)
                        for(int j=1;j<=m;j++)
                        {
                                dp[i][j]=dp[i-1][j];
                                if(j>=weight[i])
                                        dp[i][j]=max(dp[i-1][j],dp[i-1][j-weight[i]]+value[i]);
                        }
                printf("%d\n",dp[n][m]);
        }
        return 0;
}