algorithm ch15 FastWay

这是DP最基础的一个问题，刚开始学习这一块，实现了一下，不要黑我巨长的参数表，新手。

代码如下：

void FastWay(int l1[], int l2[], int e1, int e2, int x1, int x2, int t1[], int t2[], int n)
{
    int f1[] ={};
    int f2[] ={};
    int l[] = {};
    f1[] = e1 + l1[];
    f2[] = e2 + l2[];
    int lres = ;
    int fres = ;
    int offset = ;

    for(int i = ; i < n; ++i)
    {
        if((f1[i-] + l1[i]) <= (f2[i-] + t2[i-] + l1[i]))
        {
            l[i] = ;
            f1[i] = f1[i-] + l1[i];
        }
        else
        {
            f1[i] = f2[i-] + t2[i-] + l1[i];
            l[i] = ;
        }
        if(f2[i-] + l2[i] <= f1[i-] + t1[i-] + l2[i])
        {
            l[i+] = ;
            f2[i] = f2[i-] + l2[i];
        }
        else
        {
            l[i+] = ;
            f2[i] = f1[i-] + t1[i-] + l2[i];
        }
    }
    if(f1[n-] + x1 <= f2[n-] + x2)
    {
        lres = ;
        fres = f1[n-] + x1;

    }
    else
    {
        lres = ;
        fres = f2[n-] + x2;
        offset = ;
    }
    PrintWayRecurse(l, lres, n);
}
void PrintWayRecurse(int l[], int lres, int n)
{
    if(n == )
        return;
    int offset = ;

    if(lres == )
    {
        offset = ;
    }
    else
    {
        offset = ;
    }
    PrintWayRecurse(l, l[n - +offset], n- );
    cout << "line " << lres << " station " << n << endl;

}

这里是迭代输出的，当时教c语言的老湿真的是极力的反对我们用迭代，所有的迭代都可以用顺序语句实现，可是现在看算法这块好多将迭代的，感觉思维有点跟不上，困惑。

然后写了个顺序执行的：

void PrintWay(int l[], int lres, int offset, int n)
{
    if(offset == )
    {
        cout << "line " <<   << " station" << n << endl;
    }
    else
    {
        cout << "line " <<  << " station" << n << endl;
    }
    for(int i = n-; i> ; --i)
    {
        cout << "line " << l[i + offset] << " station" << i << endl;
        if(l[i+offset] == )
            offset = ;
        else
            offset = ;
    }
    cout << "test" ;
}

代码有点乱，新手。。。