算法：求行列式 C语言实现

兴趣来了，编写一个求行列式的程序，关键要解决两个问题

1）行列式列序号的全排列

2）行列式的符号

#include <iostream>
#include <vector>

using namespace std;

#define LEN 3
#define COUNT 3*2*1  // the count of premutation for length 3

int idx[LEN] = { 0,1,2};

int premus[COUNT][LEN+1]; // last elemet for store inversion number
int premus_pos = 0;

int box[LEN][LEN] = { // sample data
        { 1, 2, 3 },
        { 4, 5, 5 },
        { 2, 4, 6 } };

void arrayCopy(int src[], int dst[]) {
    for( int i=0; i<LEN; i++)
        dst[i] = src[i];
}

int inveNumber(int a[]){  // return inversion number
    int num = 0;
    for(int i=1; i<LEN; i++){
        for( int j=0; j<i; j++){
            if ( a[i] < a[j])
                num++;
        }
    }
    return num;
}


void premu( int a[], int begin, int size) { // premutation
    if ( begin == size-1){
        arrayCopy( idx, premus[premus_pos]);
        premus[premus_pos][LEN] = inveNumber(idx);

        for( int i=0; i<= LEN; i++)
            cout << " " << premus[premus_pos][i];
        cout << endl;
        premus_pos++;
        return;
    }

    premu( a, begin+1, size);
    for(int i=begin+1; i< size; i++) {
        //switch first element
        int t=a[begin];
        a[begin] = a[i];
        a[i] = t;
        //recurisive
        premu( a,begin+1, size);
        //switch back
        a[i] = a[begin];
        a[begin] = t;
    }
}

int main() {
    cout << "---go---" << endl;
    // get permeations
    premu(idx,0,LEN);
    cout << "premus_pos:" << premus_pos << endl;

    int result = 0;
    for(int i=0; i<COUNT; i++){
        int item=1;
        for( int j=0; j<LEN; j++){
            item *= box[j][premus[i][j]];
        }
        // check sign
        if ( premus[i][LEN]%2 != 0)
            item = -item;
        result += item;
    }
    cout << "result is " << result << endl;
    return 0;
}


#include <iostream>
#include <vector>

using namespace std;

#define LEN 3
#define COUNT 3*2*1

int idx[LEN] = { 0,1,2};

int premus[COUNT][LEN+1]; // last elemet for store inversion number
int premus_pos = 0;


int box[LEN][LEN] = {
        { 1, 2, 3 },
        { 4, 5, 5 },
        { 2, 4, 6 } };

/*
int box[LEN][LEN] = {
        { 1, 2 },
        { 6, 7 } };
*/
void arrayCopy(int src[], int dst[]) {
    for( int i=0; i<LEN; i++)
        dst[i] = src[i];
}

int inveNumber(int a[]){
    int num = 0;
    for(int i=1; i<LEN; i++){
        for( int j=0; j<i; j++){
            if ( a[i] < a[j])
                num++;
        }
    }
    return num;
}


void premu( int a[], int begin, int size) {
    if ( begin == size-1){
        arrayCopy( idx, premus[premus_pos]);
        premus[premus_pos][LEN] = inveNumber(idx);

        for( int i=0; i<= LEN; i++)
            cout << " " << premus[premus_pos][i];
        cout << endl;
        premus_pos++;
        return;
    }

    premu( a, begin+1, size);
    for(int i=begin+1; i< size; i++) {
        //switch first element
        int t=a[begin];
        a[begin] = a[i];
        a[i] = t;
        //recurisive
        premu( a,begin+1, size);
        //switch back
        a[i] = a[begin];
        a[begin] = t;
    }
}

int main() {
    cout << "---go---" << endl;
    // get permeations
    premu(idx,0,LEN);
    cout << "premus_pos:" << premus_pos << endl;

    int result = 0;
    for(int i=0; i<COUNT; i++){
        int item=1;
        for( int j=0; j<LEN; j++){
            item *= box[j][premus[i][j]];
        }
        // check sign
        if ( premus[i][LEN]%2 != 0)
            item = -item;
        result += item;
    }
    cout << "result is " << result << endl;
    return 0;
}


int main() {
    cout << "---go---" << endl;
    // get permeations
    premu(idx,0,LEN);
    cout << "premus_pos:" << premus_pos << endl;

    int result = 0;
    for(int i=0; i<COUNT; i++){
        int item=1;
        for( int j=0; j<LEN; j++){
            item *= box[j][premus[i][j]];
        }
        // check sign
        if ( premus[i][LEN]%2 != 0)
            item = -item;
        result += item;
    }
    cout << "result is " << result << endl;
    return 0;
}

---- 输出如下--- 可以看了列序号为  0 1 2 的全排列和对应的逆序数
---go---
 0 1 2 0
 0 2 1 1
 1 0 2 1
 1 2 0 2
 2 1 0 3
 2 0 1 2
premus_pos:6

result is 0  // 当然是 0， sample data 中有两行对应成比例