题目链接
题目分析
1、用树来表示商品分销,每个结点代表一个人,计算最后的收益,即叶节点的值
2、每增加一层,加价r%
3、结点编号0 ~ N-1
,根节点为0;上限10^5
;结果保留4
位小数
解题思路
1、树的 静态存储 (结点无权值 - 简化形式),边输入边建树;
2、先序遍历,记录当前结点深度,到达叶结点时判断、更新。
AC程序(C++)
/**************************
*@Author: 3stone
*@ACM: PAT.A1106 Lowest Price in Supply Chain
*@Time: 18/7/30
*@IDE: VSCode 2018 + clang++
***************************/
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
using namespace std;
const int maxn = 100010;
int n, min_num, min_depth;
double p, r;
//树结点
vector<int> child[maxn];
//先根遍历(DFS)
void pre_order(int root, int depth) {
if(child[root].size() == 0){ //叶节点
if(depth == min_depth) min_num++;
else if(depth < min_depth) {
min_num = 1;
min_depth = depth;
}
return;
}
for(int i = 0; i < child[root].size(); i++){
pre_order(child[root][i], depth + 1);
}
}
int main() {
while(scanf("%d %lf %lf", &n, &p, &r) != EOF) {
//初始化
for(int i = 0; i <= n; i++){
child[i].clear();
}
min_num = 0;
min_depth = maxn;
r /= 100; //r%
//获取结点信息
int temp_num, temp;
for(int i = 0; i < n; i++) {
scanf("%d", &temp_num);
if(temp_num != 0) { //非叶节点
for(int j = 0; j < temp_num; j++) {
scanf("%d", &temp);
child[i].push_back(temp);
}
}
}
//先根遍历
pre_order(0, 0); //初始层次
printf("%.4f %d\n", p * pow(r + 1, min_depth), min_num);
}//while-scanf
return 0;
}