PAT 1079 Total Sales of Supply Chain

1079 Total Sales of 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 total sales from all the 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 unit 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, then instead the total amount of the product will be given after Kj. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the total sales we can expect from all the retailers, accurate up to 1 decimal place. It is guaranteed that the number will not exceed 1010.

Sample Input:

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

Sample Output:

42.4

 这道题跟PAT 1106 很像 都是零销商……呃……PAT 1106自己好像没有整理，看心情整理的……罢了罢了

题意：求所有零售商的总销售额

把图画出来其实就是一个树，叶子结点就是零售商，有一点就是题上说 第i行第一个数为0就说明这个人是零售商 后面跟的那个数是商品的数目

方法：bfs

#include <bits/stdc++.h>
#define Max 100005
using namespace std;
struct Node{
  int layer;
  int sum; //商品数目 只有retailer有
  vector<int> child; //孩子 
}node[Max];
int n, k, v;
int leaf[Max];
double p, r;
void bfs(int rt){
  queue<int> q;
  while(!q.empty()) q.pop();
  q.push(rt);
  node[rt].layer = 0;
  while(!q.empty()){
    int u = q.front(); q.pop();
    for(int i = 0; i < node[u].child.size(); i++){
      node[node[u].child[i]].layer = node[u].layer + 1; 
      q.push(node[u].child[i]); 
    } 
  }
}
int main(){
  memset(leaf, 0, sizeof(leaf));
  cin >> n >> p >> r;
  for(int i = 0; i < n; i++){
    cin >> k;
    if(k == 0){
      cin >> node[i].sum;
      leaf[i] = 1;
    }else {
      for(int j = 0; j < k; j++){
        cin >> v;
        node[i].child.push_back(v);
      }
    }
  }
  bfs(0);
  double ans = 0, price;
//  cout << "---==" << p << endl;
  for(int i = 0; i < n; i++){
    if(leaf[i]){
      price = p * pow(1+r/100.0, node[i].layer);
//      cout << "----> " << node[i].sum << " " << node[i].layer <<" "<<price<< endl;
      ans += price * node[i].sum;
    }
  }
  cout << fixed << setprecision(1) << ans << endl;
  return 0;
}

dfs 版本

#include <bits/stdc++.h>
#define Max 100005
using namespace std;
int maxlevel=0,num=0;
struct Node {
  int sum;
  vector<int> child;
} node[Max];
double ans;
int n, k, v;
double p, r;
void dfs(int index,int level) {
  if(node[index].child.size()==0) { //叶节点
    ans+=node[index].sum*pow(1+r, level);
    return;
  }
  for(int i=0; i<node[index].child.size(); i++) 
    dfs(node[index].child[i], level+1);
}
int main() {
  cin >> n >> p >> r;
  r = r / 100.0;
  for(int i = 0; i < n; i++) {
    cin >> k;
    if(k == 0) {
      cin >> node[i].sum;
    } else {
      for(int j = 0; j < k; j++) {
        cin >> v;
        node[i].child.push_back(v);
      }
    }
  }
  dfs(0,0);
  cout << fixed << setprecision(1) << p * ans << endl;

  return 0;
}