洛谷-P3389-高斯消元

题目链接：https://www.luogu.org/problem/P3389

题意：给你方程组，如果存在多组解，输入 No solution，如果有唯一解，输出唯一解。

 比如上面这个方程组，怎么用计算机解？

高斯消元算法描述：看大佬文章吧，比我写的好，总的思路就是矩阵的行列式变化。最后把行列式变成上三角形状，盗用女装大佬的一个图

 图上最后一列是方程右边的值，前面的列是变换后的系数，可以发现，在最后一行就只有一个系数，求出最后一行的变量解后，倒数第二行也就只要求一个变量。

大佬博客地址：

https://45475.blog.luogu.org/linear-equation-group

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Main
{
    static Scanner scanner;

    static int N = 110;
    // static int N = 5;

    static double[][] a = new double[N][N];
    static double[] result = new double[N];

    static int n;

    public static void init()
    {
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                a[i][j] = 0;
            }
            result[i] = 0;
        }
    }

    public static void main(String[] args) throws IOException
    {
        InputStream in = System.in;
        //InputStream in = new FileInputStream(new File("D://1.text"));
        Reader.init(in);
        n = Reader.nextInt();
        // Reader.next();
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                a[i][j] = Reader.nextInt();
            }
        }

        // 高斯消元算法
        for (int c = 0; c < n - 1; c++)
        {
            int maxR = c;
            // 寻找最大值的行
            for (int i = maxR + 1; i < n; i++)
            {
                if (Math.abs(a[i][c]) > Math.abs(a[maxR][c]))
                {
                    maxR = i;
                }
            }
            if (maxR != c)
            {
                swapRow(maxR, c, n + 1);
                maxR = c;
            }
            dump();
            // 检查a[maxR][c] 是否等于0
            if (a[maxR][c] == 0)
            {
                break;
            }
            for (int r = c + 1; r < n; r++)
            {
                // 判断a[r][c]是否是0
                if (a[r][c] == 0)
                {
                    // 当前的列已经等于0，不用再消元
                    continue;
                }
                // 对当前行的每一列开始减数
                double val = a[maxR][c] / a[r][c];
                for (int c2 = c; c2 <= n; c2++)
                {
                    a[r][c2] = a[r][c2] * val - a[maxR][c2];
                }
            }
            dump();
        }

        // 判断是否无解
        int solution = 1;
        for (int i = 0; i < n; i++)
            if ((a[i][i] == 0 && a[i][n] != 0) || (a[i][i] == 0 && a[i][n] ==0))
            {
                solution = 0;
                break;
            }
        if (solution == 0)
        {
            System.out.println("No Solution");
        }
        else
        {
            // 求解
            int k = 0;
            for (int i = n - 1; i >= 0; i--)
            {
                double total = 0;
                for (int j = 0; j < k; j++)
                {
                    total += result[n - j - 1] * a[i][n - j - 1];
                }
                if (a[i][n - 1 - k] == 0)
                {
                    result[i] = 1;
                }
                else
                {
                    result[i] = (a[i][n] - total) / a[i][n - 1 - k];
                }

                k++;
            }
            for (int i = 0; i < n; i++)
            {
                System.out.printf("%.2f
", result[i]);
            }
        }
    }

    static void dump()
    {
//        for (int i = 0; i < n; i++)
//        {
//            for (int j = 0; j <= n; j++)
//            {
//                System.out.print(" " + a[i][j]);
//            }
//            System.out.println();
//        }
//        System.out.println();
//        System.out.println();
    }

    static void swapRow(int i, int j, int n)
    {
        for (int c = 0; c < n; c++)
        {
            double t = a[j][c];
            a[j][c] = a[i][c];
            a[i][c] = t;
        }
    }

    static class Reader
    {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /** call this method to initialize reader for InputStream */
        static void init(InputStream input)
        {
            reader = new BufferedReader(new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        /** get next word */
        static String next() throws IOException
        {
            while (!tokenizer.hasMoreTokens())
            {
                String s = reader.readLine();
                if (s == null) return s;
                tokenizer = new StringTokenizer(s);
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException
        {
            return Integer.parseInt(next());
        }

        static double nextDouble() throws IOException
        {
            return Double.parseDouble(next());
        }
    }
}