模板-->Guass消元法(求解多元一次方程组)

如果有相应的OJ题目,欢迎同学们提供相应的链接

相关链接

• 所有模板的快速链接

简单的测试

None

代码模板

/*
 * TIME COMPLEXITY:O(n^3)
 * PARAMS:
 *      a       The responding matrix of equation set.
 *      n       number of unknown factor.
 *      l       l[] mean if it is the free unit.
 *      ans     answer of unknown factor.
 */
int solve(double a[][MAXN],bool l[],double ans[],const int& n){
    int res=0,r=0;
    for(int i=0;i<n;i++)
        l[i]=false;
    for(int i=0;i<n;i++){
        for(int j=r;j<n;j++)
            if(fabs(a[j][i])>EPS){  //Be sure current value larger than EPS.
                for(int k=i;k<=n;k++)
                    swap(a[j][k],a[r][k]);
                break;
            }
        if(fabs(a[r][i])<EPS){  //The value of every cell in current column smaller than EPS or 0.
            res++;
            continue;
        }
        //elimination
        for(int j=0;j<n;j++)
            if(j!=r && fabs(a[i][j])>EPS){
                double tmp=a[j][i]/a[r][i];
                for(int k=i;k<=n;k++)
                    a[j][k]-=tmp*a[r][k];
            }
        l[i]=true,++r;
    }
    /*
     * Here we got a new matrix,like this.
     * * * * * * *
     * 0 * * * * *
     * 0 0 * * * *
     * 0 0 0 * * *
     * 0 0 0 0 * *
     * 0 0 0 0 0 *
     *
     * Got ans.
     */
    for(int i=0;i<n;i++)
        if(l[i])
            for(int j=0;j<=n;j++)
                if(fabs(a[i][j])>0)
                    ans[i]=a[j][n]/a[j][i];
    return res;
}