动态规划—北大POJ Charm Bracelet（01背包问题）

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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 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

4 61 42 63 122 7

Sample Output

题目大意：已知背包最大容纳的重量、各物品的重量和物品的价值，将物品装入背包里，使包里的总价值达到最大

题目思路：用动态规划求解，这道题是学习动态规划的最好的、也是最基础的题目

AC代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int f[14000];       //注意数组要开的足够大才可以，否则WA
int w[3500],d[3500];
int max(int a,int b) //用了求最大值
{
if(a>=b)
  return a;
else
  return b;
}
int dp(int n,int m)
{
    int i,j;
    memset(f,0,sizeof(f));
    memset(w,0,sizeof(w));
memset(d,0,sizeof(d));

    for(i=1;i<=n;i++)     //输入每个物品的重量和价值
    scanf("%d%d",&w[i],&d[i]);

    for(i=1;i<=n;i++)    //外层for循环用来控制要装入的物品的重量
    {
    for(j=m;j>=w[i];j--) //内层for循环用来控制是否要装入该物品，如果j>=w[i]就先假设把它装进去，否则无法装入
    {
    f[j]=max(f[j-w[i]]+d[i],f[j]); //假设装入物品，而且比不装时价值大时才将它装入，否则不将物品装入
    }
    }
    return f[m];      //f[m]表示背包装满时的价值，也是最大值
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF) //输入物品总数和背包的最大容量
    {
     printf("%d\n",dp(n,m));
}
return 0;
}

关键点：可以先通过二维数组进行理解，但是用二维数组的坏处是：数组开太大会超内存，太小时WA，所以最后要将它转换成一维数组

解题体会：动态规划真不好学啊，还是没有掌握到动态规划的精髓，还要继续努力啊

解题时间：2014年8月1日

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