题目：华为应用商城举办下应用得积分的活动，月末你还有40兆流量未使用，现在有4个应用可以下载，每个应用需要的流量分别为 12, 13, 23, 36，下载每个应用获得的积分为 10, 11, 20, 30，求用户可以获得的最大积分数为多少？
对应到背包问题，V=40，C={12, 13, 23, 36}, W= {10, 11, 20, 30}。
def solution1(V, C, W):
# 如果没有货物，返回0
if len(C)==0: return 0
# 不选第一个物品时的结果
ret0 = solution1(V, C[1:], W[1:])
# 如果第一个物品的占用空间大于容量V，则必不选
if C[0]>V: return ret0
# 选第一个物品时的结果
ret1 = W[0] + solution1(V-C[0], C[1:], W[1:])
# 返回选与不选之间的最大值
return max(ret0, ret1)
V = 40
C = [12, 13, 23, 36]
W = [10, 11, 20, 30]
print(solution1(V, C, W))
对应的c++代码为：
#include <iostream>
#include <vector>
using namespace std;
int solution1(int V, vector<int> C, vector<int> W){
if(C.size()==0) return 0;
int ret0 = solution1(V, vector<int>(C.begin()+1, C.end()),
vector<int>(W.begin()+1, W.end()));
if(C[0]>V) return ret0;
int ret1 = W[0]+solution1(V-C[0], vector<int>(C.begin()+1, C.end()),
vector<int>(W.begin()+1, W.end()));
return ret0>ret1?ret0:ret1;
}
int main(){
int V = 40;
vector<int> C{12, 13, 23, 36};
vector<int> W{10, 11, 20, 30};
cout << solution1(V, C, W) << endl;
return 0;
}
缺点：
- 存在大量重复计算；
- 需要大量的数组拷贝；
解决方法：解决第一个问题，通过字典记录下所有求解过的问题。
def solution2(V, C, W, record):
if len(C)==0: return 0
if (V, len(C)) in record: return record[(V, len(C))] # +++
ret0 = solution2(V, C[1:], W[1:], record);
if C[0]>V: return ret0
ret1 = W[0] + solution2(V-C[0], C[1:], W[1:], record)
record[(V, len(C))] = max(ret0, ret1) # +++
return max(ret0, ret1)
V = 40
C = [12, 13, 23, 36]
W = [10, 11, 20, 30]
print(solution2(V, C, W, dict()))
对应的C++代码：
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int solution2(int V, vector<int> C, vector<int> W, map<pair<int,int>,int>& record){
pair<int, int> situation(V, C.size()); //+++
auto p = record.find(situation); //+++
if(p!=record.end()) return p->second; //+++
if(C.size()==0) return 0;
int ret0 = solution2(V, vector<int>(C.begin()+1, C.end()),
vector<int>(W.begin()+1, W.end()),record);
if(C[0]>V) return ret0;
int ret1 = W[0]+solution2(V-C[0], vector<int>(C.begin()+1, C.end()),
vector<int>(W.begin()+1, W.end()), record);
record[situation] = ret0>ret1?ret0:ret1;//+++
return ret0>ret1?ret0:ret1;
}
int main(){
int V = 40;
vector<int> C{12, 13, 23, 36};
vector<int> W{10, 11, 20, 30};
map<pair<int,int>, int> record; //+++
cout << solution2(V, C, W, record) << endl;
return 0;
}
解决第二个问题：递归调用时需要大量的数组拷贝，解决数组拷贝问题 -- 增加一个参数证明带选择的物品。
def solution3(V, C, W, record, i):
if i>=len(C): return 0
if (V, i) in record: return record[(V, i)]
ret0 = solution3(V, C, W, record, i+1);
if C[i]>V: return ret0
ret1 = W[i] + solution3(V-C[i], C, W, record, i+1)
record[(V, i)] = max(ret0, ret1)
return max(ret0, ret1)
V = 40
C = [12, 13, 23, 36]
W = [10, 11, 20, 30]
print(solution3(V, C, W, dict(), 0))
对应的c++代码：
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int solution3(int V, vector<int>& C, vector<int>& W, map<pair<int,int>,int>& record, int i){
pair<int, int> situation(V, C.size());
auto p = record.find(situation);
if(p!=record.end()) return p->second;
if(i>=C.size()) return 0;
int ret0 = solution3(V, C, W, record, i+1);
if(C[i]>V) return ret0;
int ret1 = W[i]+solution3(V-C[i], C, W, record, i+1);
record[situation] = ret0>ret1?ret0:ret1;
return ret0>ret1?ret0:ret1;
}
int main(){
int V = 40;
vector<int> C{12, 13, 23, 36};
vector<int> W{10, 11, 20, 30};
map<pair<int,int>, int> record;
cout << solution3(V, C, W, record, 0) << endl;
return 0;
}
注意：python数组传递时默认是传引用，c++中vector传递时默认是值传递。
另一种思路：用一个 (N+1) × (V+1) 的数组A记录计算过的结果，索引(i, v) 表示如果物品为前i个，容量为v时，可以获取的最大价值。转移函数 可以写成：
注意：如果, 则
。代码为：
def solution4(V, C, W):
N = len(C)
record = [[0]*(V+1) for i in range(N+1)]
for i in range(1, N+1):
for v in range(1, V+1):
ret1 = record[i-1][v]
ret2 = 0
if v-C[i-1] >= 0:
ret2 = W[i-1]+record[i-1][v-C[i-1]]
record[i][v] = max(ret1, ret2)
return record[N][V]
V = 40
C = [12, 13, 23, 36]
W = [10, 11, 20, 30]
print(solution4(V, C, W))
对应c++代码：
#include <iostream>
#include <vector>
using namespace std;
int solution4(int V,const vector<int>& C,const vector<int>& W){
vector< vector<int> > record;
int N = C.size();
for(int i=0; i<N+1; ++i)
record.push_back(vector<int>(V+1, 0));
for(int i=1; i<N+1; ++i){
for(int v=1; v<V+1; ++v){
int ret1 = record[i-1][v];
int ret2 = 0;
if(v-C[i-1]>=0)
ret2 = W[i-1]+record[i-1][v-C[i-1]];
record[i][v] = ret1>ret2?ret1:ret2;
}
}
return record[N][V];
}
int main(){
int V = 40;
vector<int> C{12, 13, 23, 36};
vector<int> W{10, 11, 20, 30};
cout << solution4(V, C, W) << endl;
return 0;
}
上述算法的时间复杂度为：O(NV)，空间复杂度为O(NV)，下面将空间复杂度降为 O(V)。(如果要指明选出那几件物品，则时间复杂度不能降低)。
def solution5(V, C, W):
N = len(C)
record = [0]*(V+1)
for i in range(0, N):
for v in range(V, -1, -1):
ret1 = record[v]
ret2 = 0
if v-C[i] >= 0:
ret2 = W[i]+record[v-C[i]]
record[v] = max(ret1, ret2)
return record[V]
V = 40
C = [12, 13, 23, 36]
W = [10, 11, 20, 30]
print(solution5(V, C, W))
对应的c++代码为：
#include <iostream>
#include <vector>
using namespace std;
int solution5(int V,const vector<int>& C,const vector<int>& W){
vector<int> record(V+1, 0);
int N = C.size();
for(int i=0; i<N; ++i){
for(int v=V; v>=0; --v){
int ret1 = record[v];
int ret2 = 0;
if(v-C[i]>=0)
ret2 = W[i]+record[v-C[i]];
record[v] = ret1>ret2?ret1:ret2;
}
}
return record[V];
}
int main(){
int V = 40;
vector<int> C{12, 13, 23, 36};
vector<int> W{10, 11, 20, 30};
cout << solution5(V, C, W) << endl;
return 0;
}
非表格形式的解决方法适用于V特别大，而物品数量较少的情况，不可用于完全背包问题。
表格形式的解决方案适用于V较小，而物品数量较多的情况，可以用于完全背包问题。
备注：
- 第二种方法，record如果初始化为
-inf，则为更好满足容量的背包问题；
def solutionX(V, C, W):
N = len(C)
record = [-float('inf')]*(V+1)
for i in range(0, N):
for v in range(V, -1, -1):
ret1 = record[v]
ret2 = 0
if v-C[i] >= 0:
ret2 = W[i]+record[v-C[i]]
record[v] = max(ret1, ret2)
return record[V]
- 第二种方法，如果内层循环方向变化，则变为0-1背包问题；
def solutionY(V, C, W):
N = len(C)
record = [0]*(V+1)
for i in range(0, N):
for v in range(0, V+1):
ret1 = record[v]
ret2 = 0
if v-C[i] >= 0:
ret2 = W[i]+record[v-C[i]]
record[v] = max(ret1, ret2)
return record[V]
练习转移函数的理解：https://leetcode.com/problems/house-robber/
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
Example 2:
Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.
网友评论