PAT_（树的遍历）1094 The Largest Generation (25）1106 Lowest Price in Supply Chain (25分)

目录

1094 The Largest Generation (25分)

1106 Lowest Price in Supply Chain (25分)

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1094 The Largest Generation (25分)

A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.

家族树：找人口最多的那一层的节点数（和层数） ：思路很简单用BFS，记录每层个数，维护最大值

Input Specification:

Each input file contains one test case. Each case starts with two positive integers N (<100) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<N) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a family member, K (>0) is the number of his/her children, followed by a sequence of two-digit ID's of his/her children. For the sake of simplicity, let us fix the root ID to be 01. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.

Sample Input:

23 13
21 1 23
01 4 03 02 04 05
03 3 06 07 08
06 2 12 13
13 1 21
08 2 15 16
02 2 09 10
11 2 19 20
17 1 22
05 1 11
07 1 14
09 1 17
10 1 18

Sample Output:

9 4

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#define pb push_back 
using namespace std;
int n,m;
int maxn_l,level=1,maxn=-1;
vector<int> graph[105];

void bfs(){
	queue<int> qu;
	qu.push(1);
	int node,child;
	while(!qu.empty()){
		int len=qu.size();
		if(len>maxn){
			maxn=len;
			maxn_l=level;
		}
		level++;
		for(int i=0;i<len;i++){
			node=qu.front();qu.pop();
			for(int j=0;j<graph[node].size();j++){
				qu.push(graph[node][j]);
			}
		}	
	}
	
}
int main(){
	scanf("%d%d",&n,&m);
	int start,num,tmp;
	for(int i=0;i<m;i++){
		scanf("%d%d",&start,&num);
		while(num--){
			scanf("%d",&tmp);
			graph[start].pb(tmp);
		}
	}
	bfs();
	printf("%d %d",maxn,maxn_l);
	return 0;
}
	

 

 

1106 Lowest Price in Supply Chain (25分)

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:  题目描述不再重复了，已经第三遍了

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,

（supplier 0：供给2 3 5 ... ... ）显然注意到 除了Ki=0外，一共9条边，10个节点；刚好构成一棵树

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

题意：找深度最小的节点的个数，以及花销，思路就是对1090稍微改改！

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#define pb push_back 
using namespace std;
const int maxn=100005;

int n,cnt; 
double p,r;
vector<int> graph [maxn];
int lowest=0x3f3f3f3f;
int vis[maxn];
void dfs(int num,int depth){
	if(!graph[num].size()){	
		if(depth<lowest) {
			lowest=depth; cnt=1;
		}else if(lowest==depth){
			cnt++;
		}
		return ;
	} 
	int len=graph[num].size();	
	for(int i=0;i<len;i++){
		int node=graph[num][i];
		if(!vis[node]) {
			vis[node]=1;
			dfs(node,depth+1); 
		}
	}
}

int main(){
	scanf("%d%lf%lf",&n,&p,&r);
	r/=100.0;
	int tmp,num;
	for(int i=0;i<n;i++){
		scanf("%d",&tmp);
		if(tmp!=-1){
			while(tmp--){
				scanf("%d",&num);
				graph[i].pb(num);
			}
		}
	}
	vis[0]=1;
	dfs(0,0);
	double sum=p*pow(1+r,lowest);
	printf("%.4f %d",sum,cnt);
	return 0;
}