白书高斯消元模板

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白书高斯消元模板

『题目链接』白书P287

✿说明✿

这个就是直接模拟解方程的过程，当有具体的可解一次方程时，就可以用这个。也注意下数据范围，100左右的时候才可用。

『时间复杂度』O(n3)

✿CODE✿

#include <map>#include <cmath>#include <cstdio>#include <vector>#include <iostream>#include <set>#include <queue>#include <cstring>#include <algorithm>using namespace std;const double EPS = 1e-8;typedef vector<double> vec;typedef vector<vec> mat;vec gauss_jordan(const mat& A, const vec& b) {    int n = A.size();    mat B(n, vec(n + 1));    for (int i = 0; i < n; i++)        for (int j = 0; j < n; j++)            B[i][j] = A[i][j];    for (int i = 0; i < n; i++)        B[i][n] = b[i];    for (int i = 0; i < n; i++) {        int pivot = i;        for (int j = i; j < n; j++) {            if (abs(B[j][i]) > abs(B[pivot][i])) pivot = j;        }        swap(B[i], B[pivot]);        if (abs(B[i][i]) < EPS) return vec();        for (int j = i + 1; j <= n; j++) {            B[i][j] /= B[i][i];        }        for (int j = 0; j < n; j++) {            if (i != j) {                for (int k = i + 1; k <= n; k++)                    B[j][k] -= B[j][i] * B[i][k];            }        }    }    vec x(n);    for (int i = 0; i < n; i++)        x[i] = B[i][n];    return x;}int main() {    freopen("1.in", "r", stdin);    mat A{{1, -2, 3}, {4, -5, 6}, {7, -8, 10}};    vec B{6, 12, 21};    vec x = gauss_jordan(A, B);    if (x.size()) {        for (auto &i : x) {            cout << i << ' ';        } printf("\n");    }    else {        puts("Oops");    }    return 0;}

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