1106 Lowest Price in Supply Chain （25 point(s))

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 (≤10​5​​), 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:

K​i​​ ID[1] ID[2] ... ID[K​i​​]

where in the i-th line, K​i​​ 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. K​j​​ 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 10​10​​.

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

题目大意：给一颗树，树根结点权值已知，从根节点每向下一层，结点权值上升 r 个百分点。问你：最上层叶子结点的的权值和此层叶子结点的个数。

分析：

（1）题目给的是编号关系，因此采用静态存储。

（2）每向下一层上升r个百分点，可以采用DFS深度递归求出求出最靠近根节点的叶子结点的层数，并记录此层的叶子节点个数

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int N=100100;
struct node{
	double data;
	vector<int>child;
}node[N]; 
int num,mindepth=N;
void DFS(int index,int depth){
	if(node[index].child.size()==0){
		if(depth<mindepth){
			mindepth=depth;
			num=1;
		}
		else if(depth==mindepth) num++;
	}
	for(int i=0;i<node[index].child.size();i++)
		DFS(node[index].child[i],depth+1);	
}

int main(){
	double n,p,r;
	cin>>n>>p>>r;
	r/=100;
	for(int i=0;i<n;i++){
		int k,id;
		cin>>k;
		if(k!=0){
			for(int j=0;j<k;j++){
				cin>>id;
			    node[i].child.push_back(id);
			}	
		}
	}
	DFS(0,0);
	printf("%.4lf %d\n",p*pow((1+r),mindepth),num);
	return 0;
}

下面两道题是类似题：

1079 Total Sales of Supply Chain （25 point(s))

1090 Highest Price in Supply Chain （25 point(s))

 

1079题是求出所有的叶子节点的总权值，在上一题数据结构中增添了叶子重量而已（而不是原来的单位一）

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int N=100100;
struct node{
	double data;
	vector<int>child;
}node[N]; 
int n;double p,r,ans;
void DFS(int index,int depth){
	if(node[index].child.size()==0){
		ans+=node[index].data*pow(1+r,depth);
		return ;
	}
	for(int i=0;i<node[index].child.size();i++)
		DFS(node[index].child[i],depth+1);	
}

int main(){
    cin>>n>>p>>r;
	r/=100;
	for(int i=0;i<n;i++){
		double k,val,id;
		cin>>k;
		if(k!=0){
			for(int j=0;j<k;j++){
				cin>>id;
				node[i].child.push_back(id); 
			} 
		}
		if(k==0) cin>>node[i].data;
	}
	DFS(0,0);
	printf("%.1lf",ans*p);
	return 0;
}

1090题输出的是最下层的叶子节点的权值和此层的叶子结点的个数，对原有代码进行稍加修改即可。

AC代码：
 

#include<bits/stdc++.h>
using namespace std;
const int N=100100;
struct node{
	vector<int>child;
}node[N]; 
int num,maxdepth=-1;
void DFS(int index,int depth){
	if(node[index].child.size()==0){
		if(depth>maxdepth){
			maxdepth=depth;
			num=1;
		}
		else if(depth==maxdepth) num++;
	}
	for(int i=0;i<node[index].child.size();i++)
		DFS(node[index].child[i],depth+1);	
}

int main(){
	double n,p,r;
	cin>>n>>p>>r;
	r/=100;
	int root,id;
	for(int i=0;i<n;i++){
		cin>>id;
		if(id!=-1) node[id].child.push_back(i);
	    else root=i;
	}
	DFS(root,0);
	//cout<<num<<endl;
	printf("%.2lf %d\n",p*pow((1+r),maxdepth),num);
	return 0;
}