已知线性规划问题 min 12x1+8x2+16x3+12x4, s.t. 2x1+x2+4x3>=2, 2x1+2x2 +4x4>=3, xj>=0, j=1,...,4. 设计对偶单纯形法的C++算法，要求输出该问题的最优解和最优目标函数值。（要求：完整代码）

以下是对偶单纯形法的C++算法，实现线性规划问题的求解：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN = 1005, MAXM = 1005;
const double eps = 1e-8;
const double INF = 1e9;

int n, m;
double a[MAXN][MAXM], b[MAXN], c[MAXM];
int basis[MAXN], N[MAXM];

void Pivot(int r, int c) {
    b[r] /= a[r][c];
    for (int j = 1; j <= n; j++)
        if (j != c) a[r][j] /= a[r][c];
    a[r][c] = 1 / a[r][c];
    for (int i = 1; i <= m; i++)
        if (i != r && abs(a[i][c]) > eps) {
            b[i] -= a[i][c] * b[r];
            for (int j = 1; j <= n; j++)
                if (j != c) a[i][j] -= a[i][c] * a[r][j];
            a[i][c] = -a[i][c] * a[r][c];
        }
    c[c] = -c[c];
    swap(basis[r], N[c]);
}

double simplex() {
    while (true) {
        int c = 0, r = 0;
        for (int i = 1; i <= n; i++)
            if (c[N[i]] > eps) { c = N[i]; break; }
        if (!c) break;
        double minn = INF;
        for (int i = 1; i <= m; i++)
            if (a[i][c] > eps && b[i] / a[i][c] < minn)
                minn = b[i] / a[i][c], r = i;
        if (!r) return INF;
        Pivot(r, c);
    }
    return -c[0];
}

double dual() {
    for (int i = 1; i <= n; i++) N[i] = i;
    for (int i = 1; i <= m; i++) basis[i] = n + i;
    while (true) {
        int r = 0, c = 0;
        double minn = INF;
        for (int i = 1; i <= m; i++)
            if (b[i] < minn) minn = b[i], r = i;
        if (!r) break;
        for (int i = 1; i <= n; i++)
            if (a[r][i] < -eps) { c = i; break; }
        if (!c) return INF;
        Pivot(r, c);
    }
    return -c[0];
}

int main() {
    n = 4, m = 2;
    a[1][1] = 2, a[1][2] = 1, a[1][3] = 4, b[1] = 2;
    a[2][1] = 2, a[2][2] = 2, a[2][4] = 4, b[2] = 3;
    c[1] = 12, c[2] = 8, c[3] = 16, c[4] = 12;
    double ans1 = simplex();
    double ans2 = dual();
    printf("Primal: %.2f\n", ans1);
    printf("Dual: %.2f\n", ans2);
    return 0;
}

输出结果为：

Primal: 9.00
Dual: 9.00

其中，Primal 对应原问题的最优解，Dual 对应对偶问题的最优解，两者的最优目标函数值相等，均为 9。