01背包问题回溯法和动态规划

题目要求：

输入背包的容量v和物品的数量n；接下来n 行每行输入两个数字,第一个是物品质量，第二个是物品价值；

输出背包容纳物品的最大价值。

下面直接贴代码：

回溯法

 1 #include<iostream>//之前必须知道背包容量和n个物品
 2 #include<algorithm>
 3 using namespace std;
 4 class Property
 5 {
 6     public:
 7     int weight,profit;
 8     double average;
 9     friend bool operator<(Property a,Property b)
10     {
11         return a.average>b.average;
12     }
13 };
14 class Pack
15 {
16     public:
17         Pack(int Q,int n)//构造函数，初始化
18         {
19             capcity=Q;
20             number=n;
21             property=new Property[n+1];
22             bestp=cw=cp=0;
23         }
24         ~Pack()
25         {
26             delete []property;
27         }
28         void inputproperty()
29         {
30             for(int i=0;i<number;i++)
31             {
32                 cin>>property[i].weight>>property[i].profit;
33                 property[i].average=1.0*property[i].profit/property[i].weight;
34             }
35             sort(property,property+number);
36         }
37         int command()
38         {
39             int totalweight=0;
40             int totalvalue=0;
41             for(int i=0;i<number;i++)
42             {
43             totalweight+=property[i].weight;
44             totalvalue+=property[i].profit;
45             }
46             if(totalweight<capcity)return totalvalue;
47             backtrack(0);
48             return bestp;
49         }
50         bool bound(int i)
51         {
52             int currentp=cp;
53             int currentw=capcity-cw;
54             while(currentw>=property[i].weight&&i<number)
55             {
56                 currentw-=property[i].weight;
57                 currentp+=property[i].profit;
58                 i++;
59             }
60             if(i<number)
61             currentp+=1.0*property[i].profit*currentw/property[i].weight;
62             return currentp>bestp;
63         }
64         void backtrack(int i)
65         {
66             if(i>number-1)
67             {
68                 if(bestp<cp)
69                 bestp=cp;
70                 return;
71             }
72             if(cw+property[i].weight<=capcity)//此处必须用<=比较符号，不然会错
73             {
74                 cw+=property[i].weight;
75                 cp+=property[i].profit;
76                 backtrack(i+1);
77                 cw-=property[i].weight;
78                 cp-=property[i].profit;
79             }
80             if(bound(i+1));
81             backtrack(i+1);
82         }
83     private:
84     int capcity,number;
85     Property *property;
86     int cw,cp,bestp;
87 };
88 int main()
89 {
90     int n,m;
91     while(cin>>n>>m)
92     {
93         Pack object(n,m);
94         object.inputproperty();
95         cout<<object.command()<<endl;
96     }
97     return 0;
98 }

输入：

20 5
4 7
5 8
2 10
5 10
8 16

输出：
44

动态规划法

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4 int main()
 5 {
 6     int Capcity,weight;
 7     int *w,*p;
 8     int profit[200][200];
 9     while(cin>>Capcity>>weight)
10     {
11         memset(profit,0,sizeof(profit));
12         w=new int[weight+1];
13         p=new int[weight+1];
14         for(int i=1;i<=weight;i++)
15         cin>>w[i]>>p[i];//输入重量和价值
16         for(int i=1;i<=weight;i++)
17         {
18             for(int j=1;j<=Capcity;j++)
19             {
20                 if(w[i]>j)profit[i][j]=profit[i-1][j];
21                 else if(profit[i-1][j]<profit[i-1][j-w[i]]+p[i])
22                 profit[i][j]=profit[i-1][j-w[i]]+p[i];
23                 else profit[i][j]=profit[i-1][j];
24             }
25         }
26         for(int i=1;i<=weight;i++)
27         {
28         for(int j=1;j<=Capcity;j++)
29         {
30             cout<<profit[i][j]<<" ";
31         }
32         cout<<endl;
33         }
34     }
35     return 0;
36 }

01背包问题回溯法和动态规划