XDOJ省赛选拔赛第二场H题
省赛选拔，三人组队赛，队友之间的配合很重要，对于英文题目，精读题，否则对题意理解的偏差可能会导致出乎意料的损失。
那么，就来回忆一下这道题。
Title：
H. Stock Prices
Time Limit：
2000Ms
Content:
In this problem we deal with the calculation of stock prices. You need to know the following things about stock prices:
1.The ask price is the lowest price at which someone is willing to sell a share of a stock.
2.The bid price is the highest price at which someone is willing to buy a share of a stock.
3.The stock price is the price at which the last deal was established.
Whenever the bid price is greater than or equal to theask price, a deal is established. A buy order offering thebid price is matched with a sell order demanding theask price, and shares areexchanged at the rate of the ask price until either the sell order or the buy order ( or both) is fulfilled (i.e.,the buyer wants no more stocks, or the seller wants to sell no more stocks). You will be given a list of orders (either buy or sell) and you have to calculate, after each order, the currentask price,bid price andstock price.
Input:
The first line gives the number of test cases, at most 100. For each test case:
1.One line with an integer n(1<=n<=1000):the number of orders.
2.n lines of the form "order_type x shares at y" whereorder_type is either "buy" or "sell",x(1<=x<=1000) is the number of shares of a stock someone wishes to buy or to sell, andy(1<=y<=1000) is the desired price.
Output:
For each test cases, output n lines, each of the form "ai bi si", where ai, bi,si are the currentask,bid andstock prices, respectively, after thei-th order has been processed and all possible deals have taken place. Wherever a price is not defined, output "-" instead of the price.
Sample Input:
2
6
buy 10 shares at 100
sell 1 shares at 120
sell 20 shares at 110
buy 30 shares at 110
sell 10 shares at 99
buy 1 shares at 120
6
sell 10 shares at 100
buy 1 shares at 80
buy 20 shares at 90
sell 30 shares at 90
buy 10 shares at 101
sell 1 shares at 80
Sample Output:
- 100 -
120 100 -
110 100 -
120 110 110
120 100 99
- 100 120
100 - -
100 80 -
100 90 -
90 80 90
100 80 90
100 - 80
分析：这就是一道典型的模拟题，分析清楚股票的交易过程，然后一步步实现就可以了。那就需要我们注意一下题目中的几个问题
1.对于ask price和bid price， 题目中明确说明了是最低/高的价格，这就说明交易不是按顺序进行的，而是需要排序挑选，因此，需要用优先队列，这个细节又会引出第六个问题；
2.对于stock price， 题目中last可理解为是上次成交的价格，这就说明，如果本次交易不成功，那stock price应该取上一个成功交易的stock price，而非“-”。除非至今还未出现成功交易，否则不会输出“-”。对此，可以考虑用一个单独的变量存储price stock，时时维护与更新，并初始化为-1以标记未出现成功交易的情况；
3.对于交易成功的结果，经分析可以理解为普通的买卖交易情况，就是少的一方买入或卖出全部股票，并且以卖方的价格出售；
4.对于交易结束的条件，确实耐人寻味，最终将重点研究对象放在了当前喊价格的这个人，即当这个人无法再进行交易时本次交易结束，否则，他可以连续和多个人进行交易。
5.基于问题四，我们可以得出每次交易结束后，所剩要价中不存在能进行交易的组合（否则在执行该要价的过程中，该组合就会被匹配）。因此，每次只有关注该次新的要价即可，不需对内部原有要价再次研究。
6.与问题五形成对比，每次研究的都是新增要价，然而，新增要价并非一定是最低/高的，即可能该次新增要价根本满足不了（意思就是说可能进入优先队列后，再出来就并不是这个这个对象了），因此处理的价格和最终输出的价格可能并不一样。所以，就需要对这个要价必须先处理，后队列，再取出，最后输出，步骤不能少，也不能转换顺序，否则就会出现不一致的情况。
#include
#include
#include
#include
#include
#include
using namespace std;
typedef struct STOCK
{
int shares;
int price;
bool operator > (const struct STOCK &s) const
{
return price > s.price;
}
bool operator < (const struct STOCK &s) const
{
return price < s.price;
}
}Stock;
priority_queue > buy;
priority_queue , greater > sell;
int main()
{
//freopen("H.in","r",stdin);
//freopen("re.out","w",stdout);
int m,n;
scanf("%d",&m);
while(m--)
{
while(!buy.empty())
{
buy.pop();
}
while(!sell.empty())
{
sell.pop();
}
scanf("%d",&n);
int lastprice = -1;
for(int i=0;itempstock.shares)
{
laststock.shares -= tempstock.shares;
sell.push(laststock);
in = true;
break;
}
else if(laststock.shares=tempstock.price)
{
lastprice = tempstock.price;
buy.pop();
if(laststock.shares>tempstock.shares)
{
laststock.shares -= tempstock.shares;
buy.push(laststock);
in = true;
break;
}
else if(laststock.shares 可见，读题的过程必须耐心，仔细。如果题意读错，就可能花费大量时间做无用功，以此题为例，细节决定成败。
科普：命令行的fc命令可以比较两个文件的不同之处，确实也是一个不小的收获