//运行环境GCC6.4.0 C++11  非VC++
//实验一 单纯性方法
#include "cmath"
#include "cstdio"
#include "vector"
#include "algorithm"
#include "iostream"
using namespace std;int N, M;
double **A;
int *mark;bool Simplex(int ROW, int COL) {//两阶段法//mark标记进基变量mark = (int *)malloc(sizeof(int)*N);for (int i = 0; i < M; i++) mark[i] = M+i;//使得添加的变量的检验数为0for (int i = 2; i < N+2; i++) {for (int j = 0; j < COL; j++) {A[1][j] += A[i][j];}}//按照单纯形方法进行迭代double maxx = -1;int pos = 0;for (int i = 0; i < N+M; i++) {if (A[1][i] > maxx) {maxx = A[1][i]; pos = i;}}while (maxx > 0) {double minn = -1;int pos1 = 0;for (int i = 0; i < N; i++) {if (A[i+2][pos] > 0) {if (minn < 0) { pos1 = i+2; minn = A[i+2][M+N]/A[i+2][pos];}else if (A[i+2][M+N]/A[i+2][pos] < minn) {pos1 = i+2;minn = A[i+2][N+M]/A[i+2][pos];}}}if (minn == -1.0) return false;double s = A[pos1][pos];for (int i = 0; i < COL; i++) {A[pos1][i] /= s;}mark[pos1-2] = pos;for (int i = 0; i < ROW; i++) {if (i == pos1) continue;s = A[i][pos];for (int j = 0; j < COL; j++) {A[i][j] -= s*A[pos1][j];}}maxx = -1;pos = 0;for (int i = 0; i < N+M; i++) {if (A[1][i] > maxx) {maxx = A[1][i]; pos = i;}}}return true;
}
int main(int argc, char const *argv[])
{printf("请输入自变量的个数和方程组的个数:");scanf("%d%d", &M, &N);A = (double **)malloc(sizeof(double *)*(N+3)); //A为单纯形表for (int i = 0; i < N+3; i++) {*(A+i) = (double *)malloc(sizeof(double)*(M+N+2));}printf("请输入约束矩阵:\n");for (int i = 0; i < N; i++) {for (int j = 0; j < M; j++) {scanf("%lf", &A[i+2][j]);}}for (int i = 0; i < N; i++) {for (int j = 0; j < N; j++) {if (i == j) A[i+2][j+M] = 1.0;else A[i+2][j+M] = 0.0;}}printf("请输入价值向量:\n");for (int i = 0; i < M; i++) {double t; scanf("%lf", &t);A[0][i] = -t;}for (int i = 0; i < N + M; i++) {if (i < M) A[1][i] = 0.0;else A[1][i] = -1.0;}printf("请输入右端向量:\n");A[0][N+M] = A[1][N+M] = 0.0;for (int i = 0; i < N; i++) {scanf("%lf", &A[i+2][N+M]);}//向量R中保存的是基本解向量对应的值int *R = (int *)malloc(sizeof(int)*(N+M));for (int i = 0; i < N+M; i++) R[i] = -1;if (Simplex(N+2, M+N+1)) {double g = 0.0;//辅助问题的gfor (int i = 0; i < N; i++) {R[mark[i]] = i;if (mark[i] >= M) g += A[i][M+N];}// 如果min g != 0 方程无解if (g > 0) printf("该线性方程无解！\n");else {printf("利用单纯形方法得到的解为%.6lf\n", A[0][M+N]);printf("该线性规划的一个基本可行解为:\n");for (int i = 0; i < M; i++) {if (R[i] != -1) printf("%lf\n", A[R[i]+2][M+N]);else printf("%lf\n", 0.0);}}}else printf("该线性方程无解！\n");return 0;
}