动态规划基础之背包问题

01背包问题是最经典的背包问题，没有之一。

关于背包问题，我举过一个例子，有一天，阿里巴巴背着一个背包来到了山洞里，面对大量的金银财宝，他的背包却容量有限，他要如何选择呢？假如说这些财宝可以无限分割，例如是金粉，银粉，铜粉，他只要贪心地先放金，再放银，最后放铜，直到装满，那么如果这些财宝各有体积且无法切割，很显然，贪心无法解决这个问题，那么就只能使用动态规划了。

举个例题

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 128982    Accepted Submission(s): 51264

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

Sample Output

14

 

本题为01背包裸题，其状态转移方程为dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i])，dp[i][j]表示遍历到第i个物品，使用了j的容量时得到的最大价值，其值由是否将这个物品放入决定。w[i]表示第i个物品的体积，v[i]表示第i个物品的价值。

然而二维的空间实在太多，容易超出内存限制，我们又观察到第i行只由第i-1行决定，于是我们可以将第一个维度删除，状态转移方程变为dp[j]=max(dp[j],dp[j-w[i]]+v[i])。这个操作被称为滚动数组。可是我们又发现这样操作之后原式的dp[i-1][j-w[i]]+v[i]会被覆盖为dp[i][j-w[i]]+v[i]，出现了错误，所以在遍历j的时候我们从右往左遍历，这样就不会被覆盖了。

代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll dp[1005];
int w[1005];
int v[1005];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,vo;
        scanf("%d%d",&n,&vo);
        for(int i=1;i<=n;i++)scanf("%d",&v[i]);
        for(int i=1;i<=n;i++)scanf("%d",&w[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int j=vo;j>=w[i];j--){
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%lld\n",dp[vo]);
    }
    return 0;
}