c++实现单纯形法现行规划问题的求解(推荐)

在本程序中默认该现行规划问题有最优解

针对此问题：

1. #include<iostream>
2. using namespace std;
3.  
4. int check(float *sigema, int m) {
5. for (int i = 1; i <= m ; i++) {
6. if (sigema[i] > 0) {
7. return 0;
8. }
9. }
10. return 1;
11. }
12.  
13. //此程序已经化为标准型的线性规划问题中,且默认有最优解
14. int main(int argc, char* argv[])
15. {
16. //数据输入部分
17. int m, n;
18. cout << "请输入变量个数：";
19. cin >> m;
20. cout << "请输入不等式个数:";
21. cin >> n;
22. float **matrix = new float*[n + 1]; //系数矩阵
23. for (int i = 1; i <= n; i++) {
24. matrix[i] = new float[m + 2];
25. }
26. float *cj = new float[m + 1];
27. float *cB = new float[n + 1]; //基变量系数
28. int *XB = new int[n + 1]; //用来标注基变量x的下标
29. float *b = new float[n + 1];
30. float *sigema = new float[n + 1];
31. float *sita = new float[n + 1];
32. //初始化
33. for (int i = 0; i <= m; i++) {
34. cj[i] = 0;
35. }
36. for (int i = 0; i <= n; i++) {
37. cB[i] = 0;
38. XB[i] = 0;
39. b[i] = 0;
40. sigema[i] = 0;
41. sita[i] = 0;
42. }
43. cout << "请输入目标函数系数(用空格间开)：" << endl;
44. for (int i = 1; i <= m; i++) {
45. cin >> cj[i];
46. }
47. cout << "请输入各不等式的系数和常量(用空格间开)：" << endl;
48. for (int i = 1; i <= n; i++) {
49. cout << "不等式" << i << ": ";
50. for (int j = 1; j <= m + 1; j++) {
51. cin >> matrix[i][j];
52. }
53. }
54. cout << "请输入目标函数中基变量下标：" << endl;
55. for (int i = 1; i <= n; i++) {
56. cin >> XB[i];
57. cB[i] = cj[XB[i]];
58. //常量
59. b[i] = matrix[i][m + 1];
60. }
61.  
62. //计算检验数
63. for (int i = 1; i <= m; i++) {
64. sigema[i] = cj[i];
65. for (int j = 1; j <= n; j++) {
66. sigema[i] -= cB[j] * matrix[j][i];
67. }
68. }
69.  
70. while (check(sigema, m) == 0) {
71. //寻找入基变量
72. float maxn = sigema[1];
73. int sigema_xindex = 0;
74. float sigema_xcoefficient = 0;
75. for (int i = 1; i <= m; i++) {
76. if (maxn <= sigema[i]) {
77. maxn = sigema[i];
78. sigema_xindex = i;
79. sigema_xcoefficient = cj[i];
80. }
81. }
82. //计算sita
83. for (int i = 1; i <= n; i++) {
84. if (matrix[i][sigema_xindex] > 0) {
85. sita[i] = b[i] / matrix[i][sigema_xindex];
86. }
87. else {
88. sita[i] = 9999; //表示sita值为负数
89. }
90. }
91. //寻找出基变量
92. float minn = sita[1];
93. int sita_xindex = 0;
94. for (int i = 1; i <= n; i++) {
95. if (minn >= sita[i] && sita[i] > 0) {
96. minn = sita[i];
97. sita_xindex = i;
98. }
99. }
100. //入基出基变换,先入基再出基
101. //入基操作
102. for (int i = 1; i <= n; i++) {
103. if (i == sita_xindex) {
104. XB[i] = sigema_xindex;
105. cB[i] = sigema_xcoefficient;
106. break;
107. }
108. }
109. //出基计算
110. //化1
111. //cout << endl << "此处为化1的结果------" << endl;
112. float mul1 = matrix[sita_xindex][sigema_xindex];
113. for (int i = 1; i <= m; i++) {
114. matrix[sita_xindex][i] /= mul1;
115. }
116. b[sita_xindex] /= mul1;
117. //化0
118. //cout << endl << "此处为化0的结果------" << endl;
119. for (int i = 1; i <= n; i++) {
120. if (i == sita_xindex) {
121. continue;
122. }
123. float mul2 = matrix[i][sigema_xindex] / matrix[sita_xindex][sigema_xindex];
124. for (int j = 1; j <= m; j++) {
125. matrix[i][j] -= (matrix[sita_xindex][j] * mul2);
126. }
127. b[i] -= (b[sita_xindex] * mul2);
128. }
129. for (int i = 1; i <= n; i++) {
130. if (i == sita_xindex) {
131. continue;
132. }
133. }
134. for (int i = 1; i <= m; i++) {
135. sigema[i] = cj[i];
136. for (int j = 1; j <= n; j++) {
137. sigema[i] -= cB[j] * matrix[j][i];
138. }
139. }
140. }
141. float MaxZ = 0;
142. float *result = new float[m + 1];
143. for (int i = 0; i <= m; i++) {
144. result[i] = 0;
145. }
146. for (int i = 1; i <= n; i++) {
147. result[XB[i]] = b[i];
148. }
149. cout << "最优解为：X = (";
150. for (int i = 1; i < m; i++) {
151. cout << result[i] << ",";
152. }
153. cout << result[m] << ")" << endl;
154. for (int i = 1; i <= m; i++) {
155. MaxZ += result[i] * cj[i];
156. }
157. cout << "最优值为：MzxZ = " << MaxZ;
158. return 0;
159. }

程序运行结果：

总结

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原文链接：https://www.cnblogs.com/zz-1120-wtenlb/p/12616481.html