动态规划解决POJ 3624

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 weightW_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

View Code

#include<iostream>
using namespace std;
int f[12881];  //f[k]表示：当背包的容量为k时的最大价值
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int Num,TotalWeight,i,k,j,weight[12881],value[12881];
        cin>>Num>>TotalWeight;
    f[0]=0;
    for(i=0;i<Num;i++)
        cin>>weight[i]>>value[i];
    for(i=0;i<Num;i++)
        for(k=TotalWeight;k>=weight[i];k--)
        {
            f[k]=max(f[k],(f[k-weight[i]]+value[i]));
            //在max中的两个参数f[k], 和f[k-weight[i]]+value[i]都是表示在背包容量为k时的最大价值
            //f[k]是这个意思，就不用说了。
            //而f[k-weight[i]]+value[i]也表示背包容量为k时的最大价值是为什么呢？
            //首先，f[k-weight[i]]表示的是背包容量为k-weight[i]的容量，也就是说f[k-weight[i]]
            //表示的是容量还差weiht[i]才到k的价值，+walue[i]恰好弥补了差的这个价值。所以……

            //如果你对f[k]=max(f[k],(f[k-weight[i]]+value[i]));这句话不是很清楚，
            //下面的这句代码会对你有帮助
            //cout<<"i="<<i+1<<"  f["<<k<<"]="<<f[k]<<endl;
        }
    cout<<f[TotalWeight]<<endl;
    return 0;
}