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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1
位小数;
解题思路
1、树的 静态存储 ,边输入边建树;
2、先序遍历,记录当前结点深度,到达叶结点时累加即可;
AC程序(C++)
/**************************
*@Author: 3stone
*@ACM: PAT.A1079 Total Sales of 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;
double p, r, total_sales;
//树结点
struct Node{
vector<int> child;
double value;
}node[maxn];
//先根遍历(DFS)
void pre_order(int root, int depth) {
if(node[root].child.size() == 0){ //叶节点
total_sales += (p * pow(r + 1, depth) * node[root].value);
return;
}
for(int i = 0; i < node[root].child.size(); i++){
pre_order(node[root].child[i], depth + 1);
}
}
int main() {
while(scanf("%d %lf %lf", &n, &p, &r) != EOF) {
//初始化
for(int i = 0; i <= n; i++){
node[i].child.clear();
}
total_sales = 0;
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) { //叶节点
scanf("%lf", &node[i].value);
}
else { //非叶节点
for(int j = 0; j < temp_num; j++) {
scanf("%d", &temp);
node[i].child.push_back(temp);
}
}
}
//先根遍历
pre_order(0, 0); //初始层次
printf("%.1f\n", total_sales);
}//while-scanf
return 0;
}