PAT(A)1106 Lowest Price in Supply Chain (25point(s))(深度優先搜索)

Sample Input

10 1.80 1.00
3 2 3 5
1 9
1 4
1 7
0
2 6 1
1 8
0
0
0

Sample Output

1.8362 2

思路：

最快樂就是看到全AC，和前面有一道題幾乎一模一樣，就是深搜求最小深度，然後統計同深度的葉子數目。開始沒看到是同一棵樹，還以爲是森林。。。審題很重要！！！
代碼

#include <bits/stdc++.h>

using namespace std;

int dmax = 999999, p = 0;

int a[100010];

vector<int>v[100010];

void dfs(int pos, int step)
{
    if (v[pos].size() == 0)
    {
        if (step == dmax)
            p++;
        else if (step < dmax)
        {
            dmax = step;
            p = 1;
        }
        return;
    }

    for (int i = 0; i < v[pos].size(); ++i)
    {
        dfs(v[pos][i], step + 1);
    }
    return;
}

int main()
{

    int n;
    double P, r;
    cin >> n >> P >> r;

    for (int i = 0; i < n; ++i) a[i] = -1;

    for (int i = 0; i < n; ++i)
    {
        int x;
        cin >> x;
        if (x == 0)
            continue;
        else
        {
            int y;
            for (int j = 0; j < x; ++j)
            {
                cin >> y;
                a[y] = i;
                v[i].push_back(y);
            }
        }
    }
    int pos = 0;
    for (int i = 0; i < n; ++i) if (a[i] == -1) pos = i; // 找根節點

    dfs(pos, 0);

    printf("%.4lf %d", pow((1 + r / 100), dmax) * P, p);

    return 0;
}