1106. Lowest Price in Supply Chain

题目

A supply chain is a network of retailers（零售商）, distributors（经销商）, and suppliers（供应商）– everyone involved in moving a product from supplier to customer.

Starting from one root supplier, everyone on the chain buys products from one’s supplier in a price P and sell or distribute them in a price that is r% higher than P. Only the retailers will face the customers. It is assumed that each member in the supply chain has exactly one supplier except the root supplier, and there is no supply cycle.

Now given a supply chain, you are supposed to tell the lowest price a customer can expect from some retailers.

Input Specification:

Each input file contains one test case. For each case, The first line contains three positive numbers: N (<=105), the total number of the members in the supply chain (and hence their ID’s are numbered from 0 to N-1, and the root supplier’s ID is 0); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then N lines follow, each describes a distributor or retailer in the following format:

Ki ID[1] ID[2] … ID[Ki]

where in the i-th line, Ki is the total number of distributors or retailers who receive products from supplier i, and is then followed by the ID’s of these distributors or retailers. Kj being 0 means that the j-th member is a retailer. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the lowest price we can expect from some retailers, accurate up to 4 decimal places, and the number of retailers that sell at the lowest price. There must be one space between the two numbers. It is guaranteed that the all the prices will not exceed 1010.

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

基本思路

考察树的遍历。
1.和第1090. Highest Price in Supply Chain 一模一样，不再赘述。

DFS版代码

#include<cstdio>#include<cmath>#include<vector>using namespace std;const int maxn = 100010;struct node{    vector<int> child;//指针域 }Node[maxn]; int n;double p,r;//结点数,原价，增长率 //DFS实现,index为当前结点的下标，depth为当前结点的深度 int minDepth = maxn,num = 0;//结点最小深度及个数 void DFS(int index,int depth){    if(Node[index].child.size() == 0){//到达叶结点         if(depth < minDepth){            minDepth = depth;            num = 1;//重置         }else if(depth == minDepth){            num++;        }        return;     }    for(int i=0;i<Node[index].child.size();i++){        int kid = Node[index].child[i];        DFS(kid,depth+1);    }} int main(){    scanf("%d%lf%lf",&n,&p,&r);    int k,child;//叶结点的个数 ,孩子结点     for(int i=0;i<n;i++){        scanf("%d",&k);        for(int j=0;j<k;j++){            scanf("%d",&child);            Node[i].child.push_back(child);         }     }    DFS(0,0);//DFS入口     printf("%.4f %d\n",p*pow(1+r/100,minDepth),num);     return 0;} 

BFS版代码**

#include<cstdio>#include<cmath>#include<vector>#include<queue>using namespace std;const int maxn = 100010;struct node{    int layer;    vector<int> child;//指针域 }Node[maxn]; int n;double p,r;//结点数,原价，增长率 //通过BFS进行遍历 int minLayer = maxn,num = 0;//最小层次 ,最小层次的叶结点个数 void BFS(int root){    queue<int> q;    q.push(root);    Node[root].layer = 0;//定义根结点为第0层    while(!q.empty()){        int top = q.front();        q.pop();        if(Node[top].child.size() == 0){//如果是叶结点             if( Node[top].layer < minLayer){                minLayer = Node[top].layer;                num = 1;            }else if(Node[top].layer == minLayer){                num++;            }        }        for(int i=0;i<Node[top].child.size();i++){//访问当前点的孩子结点             int kid = Node[top].child[i];            q.push(kid);             Node[kid].layer = Node[top].layer + 1;        }    }     return;}int main(){    scanf("%d%lf%lf",&n,&p,&r);    int k,child;//叶结点的个数 ,孩子结点     for(int i=0;i<n;i++){        scanf("%d",&k);        for(int j=0;j<k;j++){            scanf("%d",&child);            Node[i].child.push_back(child);         }     }    BFS(0);//层次遍历入口     printf("%.4f %d\n",p*pow(1+r/100,minLayer),num);     return 0;}