2020牛客多校第一场D题 Quadratic Form 拉格朗日乘子

本文链接：https://blog.youkuaiyun.com/weixin_44282912/article/details/107383928

本文探讨了如何利用拉格朗日乘数法解决带有不等式约束条件的二次型问题，详细解析了求解过程，并通过数学公式推导出了最优解的表达式。

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Quadratic Form

题意

$X=(x_1, x_2, ..., x_n)^T$ , $A$ 为 $n \times n$ 的正定二次型， $b$ 为 $n \times 1$ 的列向量
求满足求 $X^TAX \leq 1$ ， $\left(X^Tb\right)^2$ 的最大的值

题解

带有不等式约束条件解极值问题, 使用拉格朗日乘子法
设拉格朗日函数 $L\left(X, \lambda \right) = X^Tb + \lambda \left( X^TAX - 1 \right)$
由KKT条件有
$\begin{cases} \frac{\partial L}{\partial X} = b + 2 \lambda AX=0 \\ \lambda (X^TAX - 1) = 0 \\ X^TAX - 1 \leq 0 \\ \lambda \leq 0 \\ \end{cases} \Rightarrow \begin{cases} b + 2 \lambda AX=0\\ X^TAX - 1 =0 \\ \lambda \leq 0 \\ \end{cases}$
$\lambda (X^TAX - 1) = 0$ , $\lambda=0$ 则 $b = 0$ , 因此令 $X^TAX - 1=0$
由 $\lambda AX=0$ , 得 $-\frac{1}{2 \lambda} A ^ {-1} b$
则
$\begin{array}{lcl} X^TAX & = &1 \\ (-\frac{1}{2 \lambda} A ^ {-1} b) ^ TA(-\frac{1}{2 \lambda} A ^ {-1} b) &= & 1 \\ \frac{1}{4 \lambda ^ 2}b^TA^{-1}AA^{-1}b &= &1 \\ b^TA^{-1}b &= & 4\lambda^2 \end{array}$
又
$\begin{array}{lcl} X^TAX & = &1 \\ X^T(-\frac{1}{2\lambda}b) &= & 1 \\ X^T b &= & -2\lambda \end{array}$
最终有
$\left(X^Tb\right)^2=4\lambda^2=b^TA^{-1}b$

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX = 2e2 + 10;
const ll mod = 998244353;

ll qpow(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

bool Gauss(ll a[][MAX << 1], int n) {//Gauss求逆
    for (int i = 0, r; i < n; i++) {
        r = i;
        for (int j = i + 1; j < n; j++)
            if (a[j][i] > a[r][i]) r = j;
        if (r != i) swap(a[i], a[r]);
        if (!a[i][i]) return false;//无解

        ll inv = qpow(a[i][i], mod - 2);
        for (int k = 0; k < n; k++) {
            if (k == i) continue;
            ll t = a[k][i] * inv % mod;
            for (int j = i; j < (n << 1); j++)
                a[k][j] = (a[k][j] - t * a[i][j] % mod + mod) % mod;
        }
        for (int j = 0; j < (n << 1); j++) a[i][j] = a[i][j] * inv % mod;
    }
    return true;
}

int n;
ll a[MAX][MAX << 1], b[MAX];

int main() {

    while (~scanf("%d", &n)) {
        memset(a, 0, sizeof(a));
        for (int i = 0; i < n; i++) {
            a[i][i + n] = 1;
            for (int j = 0; j < n; j++)
                scanf("%lld", &a[i][j]);
        }
        Gauss(a, n);
        for (int i = 0; i < n; i++) scanf("%lld", &b[i]);
        ll ans = 0;
        //calc b^T A b
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                ans = (ans + a[i][j + n] * b[i] % mod * b[j] % mod) % mod;
        printf("%lld\n", (ans + mod) % mod);
    }

    return 0;
}