动态规划
一.钢条切割
1 #include<iostream> 2 using namespace std; 3 4 5 6 int Cut_Rod(int A[], int length, int S[]) 7 { 8 int *r = new int[length + 1]; 9 for (int i = 0; i < 11; i++) 10 { 11 r[i] = 0; 12 } 13 r[0] = 0; 14 S[0] = 1; 15 if (length <= 0 || A == NULL || S == NULL) 16 { 17 return r[0]; 18 } 19 20 for (int i = 1; i <= length; i++) 21 { 22 for (int j = 1; j <= i; j++) 23 { 24 if (r[i] < A[j - 1] + r[i - j]) 25 { 26 r[i] = A[j - 1] + r[i - j]; 27 S[i - 1] = j; 28 } 29 } 30 } 31 return r[10]; 32 } 33 34 int main() 35 { 36 int A[] = { 1, 5, 8, 9, 10, 17, 17, 20, 24, 30}; 37 int S[10]; 38 39 cout << Cut_Rod(A, 10, S) << endl; 40 for (int i = 0; i < 10; i++) 41 { 42 cout << S[i] << " "; 43 } 44 cout << endl; 45 return 0; 46 }
二.矩阵链乘法
1 #include<iostream> 2 using namespace std; 3 4 inline int min(int a, int b) 5 { 6 if (a > b) 7 { 8 return b; 9 } 10 else 11 { 12 return a; 13 } 14 } 15 16 int Matrix_Chain_Order(int matrix[], int low, int high, int **m, int **s) 17 { 18 for (int i = low; i <= high; i++) 19 { 20 m[i][i] = 0; 21 } 22 for (int len = 2; len <= high - low + 1; len++) 23 { 24 for (int i = low; i <= high - len + 1; i++) 25 { 26 int j = i + len - 1; 27 for (int k = i; k < j; k++) 28 { 29 if (k == i) 30 { 31 m[i][j] = m[i][k] + m[k + 1][j] + matrix[i * 2 - 2] * matrix[k * 2 - 1] * matrix[j * 2 - 1]; 32 s[i][j] = k; 33 } 34 else if (m[i][j] > m[i][k] + m[k + 1][j] + matrix[i * 2 - 2] * matrix[k * 2 - 1] * matrix[j * 2 - 1]) 35 { 36 m[i][j] = m[i][k] + m[k + 1][j] + matrix[i * 2 - 2] * matrix[k * 2 - 1] * matrix[j * 2 - 1]; 37 s[i][j] = k; 38 } 39 } 40 } 41 } 42 return m[low][high]; 43 } 44 45 void Print_Chain(int **s, int low, int high) 46 { 47 if (low == high) 48 { 49 cout << "A" << low; 50 } 51 else 52 { 53 cout << "("; 54 Print_Chain(s, low, s[low][high]); 55 Print_Chain(s, s[low][high] + 1, high); 56 cout << ")"; 57 } 58 } 59 60 int main() 61 { 62 int A[] = { 30, 35, 35, 15, 15, 5, 5, 10, 10, 20, 20, 25}; 63 int **m = new int*[7]; 64 for (int i = 0; i < 7; i++) 65 { 66 m[i] = new int[7]; 67 } 68 69 int **s = new int*[7]; 70 for (int i = 0; i < 7; i++) 71 { 72 s[i] = new int[7]; 73 } 74 cout << Matrix_Chain_Order(A, 1, 6, m, s) << endl; 75 Print_Chain(s, 1, 6); 76 return 0; 77 }
三.最大子段和
1 #include<iostream> 2 using namespace std; 3 4 void MaxSubSum(int A[], int length, int sum[], int num[]) 5 { 6 sum[0] = A[0]; 7 num[0] = 0; 8 for (int i = 1; i < length; i++) 9 { 10 if (sum[i - 1] > 0) 11 { 12 sum[i] = A[i] + sum[i - 1]; 13 num[i] = num[i - 1]; 14 } 15 else 16 { 17 sum[i] = A[i]; 18 num[i] = i; 19 } 20 } 21 } 22 23 int main() 24 { 25 int A[] = { -2, 11, -4, 13, -5, -2 }; 26 int sum[6], num[6]; 27 MaxSubSum(A, 6, sum, num); 28 int max = 0; 29 for (int i = 0; i < 6; i++) 30 { 31 if (sum[max] < sum[i]) 32 { 33 max = i; 34 } 35 } 36 cout << sum[max] << endl; 37 for (int i = num[max]; i <= max; i++) 38 { 39 cout << A[i] << " "; 40 } 41 cout << endl; 42 return 0; 43 }
四.LCS
1 #include<iostream> 2 #include<string> 3 #include<stack> 4 using namespace std; 5 6 int LCS(string x, string y, int **b) 7 { 8 int i = x.length(); 9 int j = y.length(); 10 int *C = new int[i]; 11 for (int t = 0; t < i; t++) 12 { 13 C[t] = 0; 14 } 15 for (int m = 0; m < j; m++) 16 { 17 int num = 0; 18 for (int n = 0; n < i; n++) 19 { 20 int preNum = C[n]; 21 if (y[m] == x[n]) 22 { 23 C[n] = num + 1; 24 b[m][n] = 0; 25 } 26 else 27 { 28 if (n != 0) 29 { 30 if (C[n - 1] > C[n]) 31 { 32 C[n] = C[n - 1]; 33 b[m][n] = -1; 34 } 35 else 36 { 37 b[m][n] = 1; 38 } 39 } 40 else 41 { 42 b[m][n] = 1; 43 } 44 } 45 num = preNum; 46 } 47 } 48 return C[i - 1]; 49 } 50 51 void PrintLCS(string str, int i, int j, int **b) 52 { 53 stack<char> s; 54 while (j >= 0 && i >= 0) 55 { 56 if (b[j][i] == 0) 57 { 58 s.push(str[i]); 59 i--; j--; 60 continue; 61 } 62 if (b[j][i] == -1) 63 { 64 i--; 65 continue; 66 } 67 if (b[j][i] == 1) 68 { 69 j--; 70 continue; 71 } 72 } 73 while (!s.empty()) 74 { 75 cout << s.top() << " "; 76 s.pop(); 77 } 78 cout << endl; 79 } 80 81 int main() 82 { 83 string x = "BDCABA"; 84 string y = "ABCBDAB"; 85 int **b = new int*[7]; 86 for (int i = 0; i < 7; i++) 87 { 88 b[i] = new int[6]; 89 } 90 cout << LCS(x, y, b) << endl; //4 91 PrintLCS(x, 5, 6, b); // BCBA 92 return 0; 93 }
五.0-1背包
1 #include<iostream> 2 using namespace std; 3 4 inline int min(int a, int b) 5 { 6 if (a > b) 7 { 8 return b; 9 } 10 else 11 { 12 return a; 13 } 14 } 15 16 inline int max(int a, int b) 17 { 18 if (a > b) 19 { 20 return a; 21 } 22 else 23 { 24 return b; 25 } 26 } 27 28 int KnapSack(int w[], int v[], int c, int n, int**m) 29 { 30 if (w == NULL || v == NULL || c <= 0 || n <= 0 || m == NULL) 31 { 32 return 0; 33 } 34 int JMax = min(w[n - 1] - 1, c); 35 for (int j = 0; j <= JMax; j++) 36 { 37 m[n - 1][j] = 0; 38 } 39 for (int j = w[n - 1]; j <= c; j++) 40 { 41 m[n - 1][j] = v[n - 1]; 42 } 43 for (int i = n - 2; i >= 1; i--) 44 { 45 JMax = min(w[i] - 1, c); 46 for (int j = 0; j <= JMax; j++) 47 { 48 m[i][j] = m[i + 1][j]; 49 } 50 for (int j = w[i]; j <= c; j++) 51 { 52 m[i][j] = max(m[i + 1][j], m[i + 1][j - w[i]] + v[i]); 53 } 54 } 55 if (w[0] > c) 56 { 57 m[0][c] = m[1][c]; 58 } 59 else 60 { 61 m[0][c] = max(m[1][c], m[1][c - w[0]] + v[0]); 62 } 63 return m[0][c]; 64 } 65 66 void TraceBack(int w[], int c, int n, int **m) 67 { 68 int *x = new int[n]; 69 for (int i = 0; i < n - 1; i++) 70 { 71 if (m[i][c] == m[i + 1][c]) 72 { 73 x[i] = 0; 74 } 75 else 76 { 77 x[i] = 1; 78 c -= w[i]; 79 } 80 } 81 if (m[n - 1][c] != 0) 82 { 83 x[n - 1] = 1; 84 } 85 else 86 { 87 x[n - 1] = 0; 88 } 89 for (int i = 0; i < n; i++) 90 { 91 if (x[i] == 1) 92 { 93 cout << i + 1 << " "; 94 } 95 } 96 cout << endl; 97 } 98 99 int main() 100 { 101 int w[] = { 4, 5, 6 }; 102 int v[] = { 3, 4, 5 }; 103 int c = 10; 104 int **m = new int*[3]; 105 for (int i = 0; i < 3; i++) 106 { 107 m[i] = new int[11]; 108 } 109 cout << KnapSack(w, v, c, 3, m) << endl; // 8 110 TraceBack(w, c, 3, m); // 1 3 111 112 return 0; 113 }
