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[6] ={0};
    int f2[6] ={0};
    int l[12] = {0};
    f1[0] = e1 + l1[0];
    f2[0] = e2 + l2[0];
    int lres = 0;
    int fres = 0;
    int offset = 0;

    for(int i = 1; i < n; ++i)
    {
        if((f1[i-1] + l1[i]) <= (f2[i-1] + t2[i-1] + l1[i]))
        {
            l[i] = 1;
            f1[i] = f1[i-1] + l1[i];
        }
        else
        {
            f1[i] = f2[i-1] + t2[i-1] + l1[i];
            l[i] = 2;
        }
        if(f2[i-1] + l2[i] <= f1[i-1] + t1[i-1] + l2[i])
        {
            l[i+6] = 2;
            f2[i] = f2[i-1] + l2[i];
        }
        else
        {
            l[i+6] = 1;
            f2[i] = f1[i-1] + t1[i-1] + l2[i];
        }
    }
    if(f1[n-1] + x1 <= f2[n-1] + x2)
    {
        lres = 1;
        fres = f1[n-1] + x1;
        
    }
    else
    {
        lres = 2;
        fres = f2[n-1] + x2;
        offset = 6;
    }
    PrintWayRecurse(l, lres, n);
}
void PrintWayRecurse(int l[], int lres, int n)
{
    if(n == 0)
        return;
    int offset = 0;

    if(lres == 2)
    {
        offset = 6;
    }
    else
    {
        offset = 0;
    }
    PrintWayRecurse(l, l[n -1 +offset], n-1 );
    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 == 6)
    {
        cout << "line " <<  2 << " station" << n << endl;
    }
    else 
    {
        cout << "line " << 1 << " station" << n << endl;
    }
    for(int i = n-1; i> 0; --i)
    {
        cout << "line " << l[i + offset] << " station" << i << endl;
        if(l[i+offset] == 1)
            offset = 0;
        else
            offset = 6;
    }
    cout << "test" ;
}

代码有点乱，新手。。。