PAT甲级 -- 1106 Lowest Price in Supply Chain (25 分)

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. 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 lowest price a customer 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 their ID's are numbered from 0 to N−1, and the root supplier's ID is 0); P, the 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:

Ki ID[1] ID[2] ... ID[Ki]

where in the i-th line, Ki 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. Kj being 0 means that the j-th member is a retailer. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the lowest price we can expect from some retailers, accurate up to 4 decimal places, and the number of retailers that sell at the lowest price. There must be one space between the two numbers. It is guaranteed that the all the prices will not exceed 1010.

Sample Input:

10 1.80 1.00
3 2 3 5
1 9
1 4
1 7
0
2 6 1
1 8
0
0
0

Sample Output:

1.8362 2

思路：

1. 建立结构体，price记录价格，childNum记录孩子个数，child记录孩子标号

2. 全局变量minPrice记录最小价格，cnt记录最小价格的个数

2. 进行输入

3. 对树进行dfs，先计算当前的价格，判断孩子个数为0，若价格低于最小价格，进行更新，cnt = 1 ！！！

若价格相等，则只更新cnt！

#include <iostream>
#include <vector>
using namespace std;
struct node{double price;int childNum;vector<int> child;
};
vector<node> supply;
int N;
double r; //N个chain上的人，r%的比率
double P; //P为最初的价格
double minPrice = 100000000000;
int cnt = 0;
void dfs(int root, double p){supply[root].price = p;if(supply[root].childNum == 0){  //到达叶子结点if(minPrice > p){cnt = 1;minPrice = p;}else if(minPrice == p){cnt++;}}for (int i = 0; i < supply[root].child.size(); i++){dfs(supply[root].child[i], p * (100+r)/100.0);}
}
int main(){scanf("%d %lf %lf", &N, &P, &r);supply.resize(N);for (int i = 0; i < N; i++){int k;scanf("%d", &k);if(k == 0) {supply[i].childNum = k;}else{supply[i].childNum = k;supply[i].child.resize(k);}for(int j = 0; j < k; j++){scanf("%d", &supply[i].child[j]);}}dfs(0,P);printf("%.4f %d", minPrice, cnt);return 0;
}

柳神代码：

等我吃过饭回来学习~~

补思路：

1. 没有用结构体，而是直接用了vector数组，v[i]表示i的孩子

2. dfs时有一个提前返回的判断，若当前的深度比最小深度大，直接返回

3.dfs统计的是最小深度，直接在最后进行计算。

#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;vector<int> v[100005];  //v[i]表示第i个的孩子结点int mindepth = 99999999, minnum = 1;void dfs(int index, int depth) {if(mindepth < depth)return ;if(v[index].size() == 0) {  //没有孩子if(mindepth == depth)minnum++;else if(mindepth > depth) {mindepth = depth;minnum = 1;}}for(int i = 0; i < v[index].size(); i++)dfs(v[index][i], depth + 1);
}int main() {int n, k, c;double p, r;scanf("%d %lf %lf", &n, &p, &r);for(int i = 0; i < n; i++) {scanf("%d", &k);for(int j = 0; j < k; j++) {scanf("%d", &c);v[i].push_back(c);}}dfs(0, 0);printf("%.4f %d", p * pow(1 + r/100, mindepth), minnum);return 0;
}