1079 Total Sales of Supply Chain (25point(s)) - C語言 PAT 甲級

1079 Total Sales of Supply Chain (25point(s))

A supply chain is a network of retailers（零售商）, distributors（經銷商）, and suppliers（供應商）-- 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:

K​i​​ ID[1] ID[2] … ID[K_i]

where in the i-th line, K​i​​ 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. K_j being 0 means that the j-th member is a retailer, then instead the total amount of the product will be given after K_j. 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

題目大意：

輸入一顆樹 N 個節點，根節點爲 0 且貨物價格爲 P，每向下走一層，貨物價格會增加 r%，

接下來輸入 N 個節點的 K 個孩子，若 K 爲 0 則表示葉節點，並給出每個葉節點給出貨物量，

求所有葉節點的售價之和

設計思路：

價格僅和葉節點層數有關，所以利用並查集的思想，記錄每個節點的父節點，以遞推出葉節點的層數

• level[v] = get_level(father[v]) + 1，要用數組 level 記錄層級信息，不記錄的話每次都查詢到根節點，會超時
• 根節點 level[0] = 0

編譯器：C (gcc)

#include <stdio.h>
#include <math.h>

double sum = 0.0, p, r;
int father[100010], level[100010], retailer[100010];

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

int main(void)
{
        int n, k, c;
        int i, j;

        scanf("%d%lf%lf", &n, &p, &r);
        r = r / 100.0;
        for (i = 0; i < n; i++) {
                scanf("%d", &k);
                if (k == 0) {
                        scanf("%d", &retailer[i]);
                } else {
                        for (j = 0; j < k; j++) {
                                scanf("%d", &c);
                                father[c] = i;
                        }
                }
        }
        for (i = 0; i < n; i++) {
                if (retailer[i] > 0) {
                        sum += retailer[i] * pow(1 + r, get_level(i));
                }
        }
        printf("%.1lf", sum * p);
        return 0;
}