dp--01背包--Charm Bracelet

Charm Bracelet

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

4 6
1 4
2 6
3 12
2 7

Sample Output

23

首先我们要明白i，j，dp[i][j]都分别代表着什么

i:表示从1~i个物品里取

j:表示此时的最大容积

dp[i][j]:表示在1~i个物品里选取，每个物品只选择一次（限于01背包），且不超过j的容积，问所能得到的最大价值

得出一个转移方程f[i][j] = max(f[i][j], f[i - 1][j - c[i]] + w[i])

即在选取这个物品和不选这个物品中选择一个最大的

最后我们要输出的就是dp[n][m]表示在n个物品中，容积不超出m的最大价值

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int dp[4000][12000];
int w[1000],c[1000];
int main()
{
    int n,m;
    scanf ("%d%d",&n,&m);
    for (int i =1 ;i<= n;i++)
    {
        scanf ("%d%d",&w[i],&c[i]);
    }
    for (int i = 1;i <= n;i++)
    {
        for (int j = 0;j <= m;j++)
        {
            dp[i][j] = dp[i-1][j];
            if (j-w[i]>=0)
            {
                dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+c[i]);
                
            }
        
        }    
    //    cout<<dp[i][m]<<endl;
    }
    cout<<dp[n][m];
    return 0;
}

先考虑上面讲的基本思路如何实现，肯定是有一个主循环i=1..N，每次算出来二维数组f[0..V]的所有值。那么，如果只用一个数组f[0..V]，能不能保证第i次循环结束后f[v]中表示的就是我们定义的状态f[v]呢？f[v]是由f[v]和f[v-c]两个子问题递推而来，能否保证在推f[v]时（也即在第i次主循环中推f[v]时）能够得到f[v]和f[v-c]的值呢？事实上，这要求在每次主循环中我们以v=V..0的顺序推f[v]，这样才能保证推f[v]时f[v-c]保存的是状态f[v-c]的值。伪代码如下：

for i=1..N

for v=V..0

f[v]=max{f[v],f[c]+w};

其中的f[v]=max{f[v],f[c]}一句恰就相当于我们的转移方程f[v]=max{f[v],f[c]}，因为现在的f[c]就相当于原来的f[c]。如果将v的循环顺序从上面的逆序改成顺序的话，那么则成了f[v]由f[c]推知，与本题意不符

#include<iostream>
using namespace std;
int f[20000];
int main()
{
    int  n,v;
    int w[20000], c[20000];
    cin >> n >> v;
    for(int i = 1; i <= n; ++i)
        cin >> w[i] >>c[i];
    for(int i = 1; i <= n; ++i){
        for (int j=v ;j>=w[i] ; j--){
                f[j] = max(f[j], f[j - w[i]] + c[i]);
        }
    }
        cout << f[v]<<endl;
    return 0;
}