1079. Total Sales of Supply Chain (25)

1.使用数的结构存储供应商和零售商

2.采用层次遍历（BFS）进行遍历

3.需要使用double进行存储，一开始发现下面几个耗时大的测试点不通过，所以考虑是精度不够或者溢出的原因，于是改为double类型，然后就通过了



AC代码：

//#include<string>
//#include <iomanip>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;

struct Node{
vector<int> list;
bool isRetailer;
int productCount;
Node() :list(0), isRetailer(false), productCount(0){};
};
int main(void)
{
int n;
double rootPrice, higher;//需要使用double
cin >> n >> rootPrice >> higher;
vector<Node> chain(n);
for (int i = 0; i < n; i++)
{
int tmp;
cin >> tmp;
if (tmp == 0)
{//零售商
chain[i].isRetailer = true;
cin >> chain[i].productCount;
}
else
{//供应商
for (int j = 0; j < tmp; j++)
{//输入下级供应商
int nextLevel;
cin >> nextLevel;
chain[i].list.push_back(nextLevel);
}
}
}

queue<int> q;
q.push(0);
int count1 = 1, count2 = 0;
double price = rootPrice;
double totalSales = 0;
//进行层次遍历
while (!q.empty())
{
for (int i = 0; i < count1; i++)
{
int head = q.front(); q.pop();
if (chain[head].isRetailer)
totalSales += chain[head].productCount*price;
else
{
for (int j = 0; j < chain[head].list.size(); j++)
{
q.push(chain[head].list[j]);
count2++;
}
}
}
count1 = count2;
count2 = 0;
price *= (1 + higher / 100.0);
}
printf("%.1lf\n", totalSales);

return 0;
}