1090. Highest Price in Supply Chain (25)（dfs）

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本文介绍了一种算法，用于计算给定供应链中零售商能够达到的最高价格及其数量。通过构建图并采用深度优先搜索（DFS），该算法可以高效地找出最长路径及对应的零售商数目。

1090. Highest Price in Supply Chain (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<=10⁵), the total number of the members in the supply chain (and hence they are numbered from 0 to N-1); 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 S_i is the index of the supplier for the i-th member. S_root for the root supplier is defined to be -1. 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 10¹⁰.

Sample Input:

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

Sample Output:

1.85 2

一开始直接用循环遍历会超时

根据给定的数组反过来，用vector邻接表建立一个图（树），然后dfs，到底部判断最深的

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
int n,num,maxnum;
double p,r;
int pre[100100];
int vis[100100];
vector<int>ans;
vector<int>G[100100];
int root;
bool cmp(int a,int b){
    return a > b;
}
void dfs(int s){
    int i;
    if(!G[s].size()){
        if(num > maxnum){
            maxnum = num;
            ans.clear();
            ans.push_back(num);
        }
        else if(num == maxnum){
            ans.push_back(num);
        }
        return;
    }
    for(i = 0; i < G[s].size(); i++){
        if(!vis[G[s][i]]){
            vis[G[s][i]] = 1;
            num++;
            dfs(G[s][i]);
            vis[G[s][i]] = 0;
            num--;
        }
    }
}
int main(){
    cin >> n >> p >> r;
    r /= 100.0;
    r += 1.0;
    memset(vis,0,sizeof(vis));
    for(int i = 0; i < n; i++){
        cin >> pre[i];
        if(pre[i] == -1) root = i;
    }
    for(int i = 0; i < n; i++){
        if(pre[i] == -1) continue;
        G[pre[i]].push_back(i);
    }
    vis[root] = 1;
    num = 0;
    maxnum = 0;
    dfs(root);
    printf("%.2f %d\n",p*pow(r,ans[0]),ans.size());
    return 0;
}