PAT 甲级 刷题日记|A 1090 Highest Price

单词积累

increment 增量 增加

题目

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 Si 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结尾无空行

Sample Output:

1.85 2结尾无空行

思路

这是一个由零售商、经销商与供应商构成的一般树。需要找到最大深度，以及该深度存在的叶节点数。

自己一开始对一般树的掌握不很好，想到了用类似并查集的方法去做，这个缺点是每次都是从叶节点处开始查找，如果出现下图所示的极端情况时，时间复杂的是O(n 方)，最后一个样例超时。

1628586481263.png

参考的做法是，用向量存储孩子，这棵树就可以表示为数组向量。数组中的第i个元素，就是以i为父节点的所有节点构成的向量。最大深度只需从根节点处开始遍历。

代码1

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100001;
int num[maxn];
int father[maxn];
int height[maxn];
int N;
double P, r;

void inital() {
    for (int i = 0; i < N; i++) {
        father[i] = i;
        height[i] = 0;
    }
}

int findfather(int x) {
    int times = 0;
    while (father[x] != -1) {
        x = father[x];
        times++;
    }
    return times;
}

void Union(int a, int b) {
    father[a] = b;
    if (b != -1) height[b] = 1;
}


int main() {
    cin>>N>>P>>r;
    inital();
    for (int i = 0; i < N; i++) {
        scanf("%d", &num[i]);
        Union(i, num[i]);
    }
    int res = 0;
    int clu = 0;
    for (int i = 0; i < N; i++) {
        if(height[i] == 0) {
            if(findfather(i) == res) clu++;
            else if(findfather(i) > res) {
                res = findfather(i);
                clu = 1;
            }
        }
    }
    double price = P * pow(1 + r * 0.01, res);
    printf("%.2f %d", price, clu);
    
} 

代码2（BFS）

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100001;
vector<int> num[maxn];
int N;
double P, r;
int maxdepth = 0;
int clus = 0;


void BFS(int root, int depth) {
    if (num[root].size() == 0) {
        
        if (depth == maxdepth) {
            clus++;
        } else if (depth > maxdepth) {
            clus = 1;
            maxdepth = depth;
        }
        return;
    }
    
    for (int i = 0; i <num[root].size(); i++) {
        BFS(num[root][i], depth + 1);
    }
}

int main() {
    cin>>N>>P>>r;
    int root = -1;
    int temp;
    for (int i  = 0; i < N; i++) {
        cin>>temp;
        if (temp == -1) root = i;
        else {
            num[temp].push_back(i);
        }
    }
    if (root != -1) BFS(root, 0);
    double price = P * pow(1 + r * 0.01, maxdepth);
    printf("%.2f %d", price, clus);
}