动态规划——背包问题变形

1 #include <iostream> 2 using namespace std; 3 #include <sys/time.h> 4 5 struct timeval t1; 6 struct timeval t2; 7 8 int n ; 9 int *a; 10 int **c; 11 int min_sub; 12 int dp(int sum,int t,int m) 13 { 14 if(t==0) 15 { 16 if( a[t] <= m) 17 return a[t]; 18 else 19 return 0; 20 } 21 int tmp1,tmp2; 22 if(m>=a[t]) 23 { 24 tmp2 = dp(sum,t-1,m-a[t]) + a[t]; 25 tmp1 = dp(sum,t-1,m); 26 if(tmp1 > tmp2) 27 return tmp1; 28 else 29 return tmp2; 30 } 31 else 32 { 33 return dp(sum,t-1,m); 34 } 35 } 36 int main() 37 { 38 gettimeofday(&t1,NULL); 39 cin >> n; 40 a = new int[n]; 41 int i = 0; 42 int j = 0; 43 for(i = 0;i<n;i++) 44 cin>>a[i]; 45 int sum = 0; 46 for(i=0;i<n;i++) 47 sum+=a[i]; 48 cout << "sum=" << sum << endl; 49 c= new int*[n]; 50 for(i=0;i<n;i++) 51 { 52 c[i] = new int[sum/2+1]; 53 for(j = 0;j<sum/2+1;j++) 54 c[i][j] = -1; 55 } 56 cout << dp(sum/2,n-1,sum/2)<<endl; 57 gettimeofday(&t2,NULL); 58 cout<< t2.tv_sec*1000+t2.tv_usec-(t1.tv_sec*1000+t1.tv_usec)<<endl; 59 } 

1 #include <iostream> 2 using namespace std; 3 #include <sys/time.h> 4 5 struct timeval t1,t2; 6 int n ; 7 int *a; 8 int **c; 9 int min_sub; 10 int dp(int sum,int t,int m) 11 { 12 if(c[t][m] !=-1) 13 return c[t][m]; 14 if(t==0) 15 { 16 if( a[t] <= m) 17 return a[t]; 18 else 19 return 0; 20 } 21 int tmp1,tmp2; 22 if(m>=a[t]) 23 { 24 tmp2 = dp(sum,t-1,m-a[t]) + a[t]; 25 tmp1 = dp(sum,t-1,m); 26 if(tmp1 > tmp2) 27 c[t][m]=tmp1; 28 else 29 c[t][m] = tmp2; 30 return c[t][m]; 31 } 32 else 33 { 34 c[t][m] = dp(sum,t-1,m); 35 return c[t][m]; 36 } 37 } 38 int main() 39 { 40 gettimeofday(&t1,NULL); 41 cin >> n; 42 a = new int[n]; 43 int i = 0; 44 int j = 0; 45 for(i = 0;i<n;i++) 46 cin>>a[i]; 47 int sum = 0; 48 for(i=0;i<n;i++) 49 sum+=a[i]; 50 cout << "sum=" << sum << endl; 51 c= new int*[n]; 52 for(i=0;i<n;i++) 53 { 54 c[i] = new int[sum/2+1]; 55 for(j = 0;j<sum/2+1;j++) 56 c[i][j] = -1; 57 } 58 cout << dp(sum/2,n-1,sum/2)<<endl; 59 gettimeofday(&t2,NULL); 60 cout<< (t2.tv_sec*1000+t2.tv_usec)-(t1.tv_sec*1000+t1.tv_usec)<<endl; 61 }  1 #include <iostream> 2 using namespace std; 3 #include <sys/time.h> 4 5 struct timeval t1,t2; 6 int n ; 7 int *a; 8 int **c; 9 int min_sub; 10 int dp(int sum) 11 { 12 int i = 0; 13 int j = 0; 14 for(i =1;i<n+1;i++) 15 for(j = 1;j<=sum;j++) 16 { 17 if(a[i] >j) 18 { 19 c[i][j] = c[i-1][j]; 20 } 21 else 22 { 23 int tmp1 = c[i-1][j]; 24 int tmp2 = c[i-1][j-a[i]] +a[i]; 25 c[i][j] = tmp1>tmp2?tmp1:tmp2; 26 } 27 } 28 return c[n][sum]; 29 } 30 int main() 31 { 32 gettimeofday(&t1,NULL); 33 cin >> n; 34 a = new int[n]; 35 int i = 0; 36 int j = 0; 37 for(i = 0;i<n;i++) 38 cin>>a[i]; 39 int sum = 0; 40 for(i=0;i<n;i++) 41 sum+=a[i]; 42 cout << "sum=" << sum << endl; 43 c= new int*[n+1]; 44 for(i=0;i<n+1;i++) 45 { 46 c[i] = new int[sum/2+1]; 47 for(j = 0;j<sum/2+1;j++) 48 c[i][j] = 0; 49 } 50 cout << dp(sum/2)<<endl; 51 gettimeofday(&t2,NULL); 52 cout<< (t2.tv_sec*1000+t2.tv_usec)-(t1.tv_sec*1000+t1.tv_usec)<<endl; 53 }  给定一个正整数的集合A={a1,a2,….,an},是否可以将其分割成两个子集合，使两个子集合的数加起来的和相等。  
  输出：例：A   =   {   1,   3,   8,   4,   10}  
  可以分割：{1,   8,   4}   及   {3,   10}

可以转化成背包问题：

1，首先算出所有元素的和sum。

2，计算不超过sum/2但最接近sum/2的元素的选择。（背包容量为suj/2）

上面给出的DP和递归的实现：

（1）迭代递归：用时104529 us

（2）记忆化递归（记忆化DP）：用时318601 us

（3）直接递归：用时634060 us