zoukankan      html  css  js  c++  java
  • PAT 甲级 1090 Highest Price in Supply Chain

    https://pintia.cn/problem-sets/994805342720868352/problems/994805376476626944

    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. 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 highest price we 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 (≤), the total number of the members in the supply chain (and hence they are numbered from 0 to N1); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then the next line contains N numbers, each number Si​​ is the index of the supplier for the i-th member. Sroot​​ for the root supplier is defined to be −. All the numbers in a line are separated by a space.

    Output Specification:

    For each test case, print in one line the highest price we can expect from some retailers, accurate up to 2 decimal places, and the number of retailers that sell at the highest price. There must be one space between the two numbers. It is guaranteed that the price will not exceed 1.

    Sample Input:

    9 1.80 1.00
    1 5 4 4 -1 4 5 3 6
    

    Sample Output:

    1.85 2

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1e5 + 10;
    int N, root, depth = INT_MIN, cnt = 0;
    double P, r;
    int vis[maxn];
    vector<int> v[maxn];
    
    void dfs(int st, int step) {
        if(v[st].size() == 0) {
            if(step > depth) {
                depth = step;
                cnt = 1;
            } else if(depth == step) cnt ++;
        }
    
        vis[st] = 1;
        for(int i = 0; i < v[st].size(); i ++) {
            if(vis[v[st][i]] == 0)
                dfs(v[st][i], step + 1);
        }
    }
    
    int main() {
        scanf("%d%lf%lf", &N, &P, &r);
        for(int i = 0; i < N; i ++) {
            int x;
            scanf("%d", &x);
            if(x == -1) root = i;
            else v[x].push_back(i);
        }
    
        memset(vis, 0, sizeof(vis));
        dfs(root, 0);
    
        double R = (r / 100 * 1.0) + 1;
        for(int i = 0; i < depth; i ++)
            P *= R;
        printf("%.2lf %d
    ", P, cnt);
        return 0;
    }
    

      dfs 今年的考试如果都是这样的题目就好了  

  • 相关阅读:
    索引与慢查询优化
    视图 触发器 事物 储存过程 内置函数 流程控制
    多表查询
    having distinct 正则 limit order by 排序
    Mysql基本查询语句及方法
    Python基础之列表内置方法
    Python基础之流程控制while循环
    Python基础之格式化输出的三种方式
    计算机基础之编程与编程语言
    计算机组成
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10383012.html
Copyright © 2011-2022 走看看