浙大 PAT 甲级 1079 Total Sales of Supply Chain 最简单的深度优先搜索 DFS

这几乎是PAT甲级练习题中最简单的一道DFS了…无需赘述，贴上代码供参考就好了。

#include<stdio.h>
#include<vector>
#include<map>
#include<math.h>
using namespace std;

vector<int> edge[100000];
map<int, int> amount;
double P, r, sales = 0.0;

void DFS(int layer,int x)
{
    if (edge[x].empty() == true)
    {
        sales += pow(r, layer)*P*amount[x];
    }
    else
    {
        for (int i = 0; i < edge[x].size(); i++)
        {
            DFS(layer + 1, edge[x][i]);
        }
    }
}

int main()
{
    int N;
    scanf("%d%lf%lf", &N, &P, &r);
    r = 1 + r / 100;
    for (int i = 0; i < N; i++)
    {
        int k;
        scanf("%d",&k);
        if (k == 0)
        {
            int tmp;
            scanf("%d", &tmp);
            amount[i] = tmp;
        }
        else
        {
            for (int j = 0; j < k; j++)
            {
                int tmp;
                scanf("%d", &tmp);
                edge[i].push_back(tmp);
            }
        }
    }
    DFS(0, 0);
    printf("%.1f", round(sales * 10) / 10.0);
    return 0;
}

标签：tmp,PAT,Chain,1079,int,scanf,DFS,edge,include
来源： https://blog.csdn.net/qq_39115541/article/details/100176836