数值分析实验报告 Lab5 LU分解算法

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数值分析实验报告 Lab5 LU分解算法

#include<stdio.h>
#include<iostream>
#include<math.h>

#define MAX_SIZE 100 /* 矩阵最大维数 */
#define ZERO 0.000000001 /* 当一个正数小于ZERO就认为该数是0 */
using namespace std;
/*
 LU方法计算 Ax=b
 */
bool Direct( int n, double a[][MAX_SIZE], double b[] ){
    /*初始化矩阵，向量*/
    double L[n][n];
    double U[n][n];
    for(int i = 0;i < n;i++){
        for(int j = 0;j < n;j++){
            L[i][j] = 0;
        if(i == j)L[i][j] = 1;
            U[i][j] = 0;
        }
    }
    double y[n];
    for(int i = 0;i < n;i++){
        y[i] = 0;
    }
    for(int k = 0;k < n;k++){
        for(int j1 = k;j1 < n;j1++){
            double sum_U = 0;
            for(int x1 = 0;x1 < k;x1++){
                sum_U += L[k][x1]*U[x1][j1];
            }
            U[k][j1] = a[k][j1] - sum_U;
            //cout<<U[k][j1]<<endl;
            if((j1==k)&&(fabs(U[k][j1])<ZERO)){return 0;}
        }
        //cout<<endl;
        for(int y2 = k+1;y2 < n;y2++){
            double sum_L = 0;
            for(int x2 = 0;x2 < k;x2++){
                sum_L += L[y2][x2]*U[x2][k];
            }
            L[y2][k] = (a[y2][k] - sum_L)/U[k][k];
            //cout<<L[y2][k]<<endl;
        }
        //cout<<endl;  
    }  
    //计算Ly=b中y的值
    for(int m1 = 0;m1 < n;m1++){
        double tmp1 = 0;
        for(int t1 = 0;t1 < m1;t1++){
            tmp1 += y[t1]*L[m1][t1];
        }
        y[m1] = b[m1] - tmp1;
        //cout<<y[m1]<<endl;
    }
    //计算Ux=y中x的值
    b[n-1] = y[n-1]/U[n-1][n-1];
    //cout<<b[n-1]<<endl;
    for(int m2 = n-2;m2 >= 0;m2--){
        double tmp2 = 0;
        for(int t2 = m2+1;t2 < n;t2++){
            tmp2 += U[m2][t2]*b[t2];
        }
        b[m2] = (y[m2] - tmp2)/U[m2][m2];
    }
    return b;
}
int main(){
    int n, i, j;
    double a[MAX_SIZE][MAX_SIZE], b[MAX_SIZE];
    freopen("input.txt", "r", stdin);
    
    while ( scanf("%d", &n) != EOF ) { /* 读取裁判测试用例 */
        for ( i=0; i<n; i++ ) {
            for ( j=0; j<n; j++ )
                scanf("%lf", &a[i][j]);
            scanf("%lf", &b[i]);
        }
        /*--- 输出直接法的解 ---*/
        if ( Direct(n, a, b) ) {
            printf("Result of direct method:\n");
            for ( j=0; j<n; j++ )
                printf("%.8lf\n", b[j]);
        }
        else
            printf("Doolittle factorization failed.\n");
        printf("\n");
    }
    //system("pause");
    fclose(stdin);  
    return 0;
}