1090 Highest Price in Supply Chain (25point(s)) - C语言 PAT 甲级

1090 Highest Price in Supply Chain (25point(s))

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

题目大意：

输入一颗树 N 个节点，根节点为 0 且货物价格为 P，每向下走一层，货物价格会增加 r%，

接下来输入 N 个节点的父节点，-1 则表示根节点

求所有叶节点的价格最大值和最大值的个数

设计思路：

1079 Total Sales of Supply Chain (25point(s)) - C语言 PAT 甲级
1106 Lowest Price in Supply Chain (25point(s)) - C语言 PAT 甲级

价格仅和叶节点层数有关，所以利用并查集的思想，记录每个节点的父节点，以递推出叶节点的层数

• level[v] = get_level(father[v]) + 1，要用数组 level 记录层级信息，不记录的话每次都查询到根节点，会超时
• 根节点 level[0] = 0
• 注意根节点是 -1 表示的，查询根节点 0 的层数时会询问 -1 节点，此时把 -1 节点看作根节点的父节点，返回 -1

编译器：C (gcc)

#include 
#include 

double p, r;
int father[100010], level[100010], root[100010];

int get_level(int v)
{
        if (v == -1)
                return -1;
        if (level[v] > 0)
                return level[v];
        level[v] = get_level(father[v]) + 1;
        return level[v];
}

int main(void)
{
        int n;
        int f, max = 0, cnt = 0;
        int i, temp;

        scanf("%d%lf%lf", &n, &p, &r);
        for (i = 0; i < n; i++)
                father[i] = -2;
        r = r / 100.0;
        for (i = 0; i < n; i++) {
                scanf("%d", &f);
                father[i] = f;
                root[f] = 1;
        }
        for (i = 0; i < n; i++) {
                if (root[i] == 0) {
                        temp = get_level(i);
                        if (temp > max) {
                                max = temp;
                                cnt = 1;
                        } else if (temp == max) {
                                cnt++;
                        }
                }
        }
        printf("%.2lf %d", pow(1 + r, max) * p, cnt);
        return 0;
}