[刷题]算法竞赛入门经典(第2版) 5-14/UVa1598 - Exchange

题意：模拟买卖，当出售价bid等于或低于出售价ask，则交易。

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代码：（Accepted，0.330s）

//UVa1598 - Exchange
//Accepted 0.330s
//#define _XIENAOBAN_
#include<functional>
#include<algorithm>
#include<iostream>
#include<utility>
#include<vector>
#include<queue>
#include<map>
#include<set>
using namespace std;

struct INFO { char type; int size, price; }for_push_back;
int N, T(0);
char type[20];
vector<INFO> LIST;                      //key: buy/ask id,  value: type, size
map<int, set<int>, greater<int> > BUY;  //key: bid price,   value: id
map<int, set<int>, less<int> > SELL;    //key: ask price,   value: id
map<int, int> BUY_VAL;                  //key: bid price,   value: size
map<int, int> SELL_VAL;                 //key: ask price,   value: size

inline int sum(set<int>& now) {
    int re(0);
    for (const auto& r : now) re += LIST[r].size;
    return re;
}

void trade(bool flag) {
    int bid_size, bid_price, ask_size, ask_price;
    while (true) {
        auto bid(BUY.begin()), ask(SELL.begin());
        if (bid == BUY.end()) bid_size = 0, bid_price = 0;
        else bid_size = BUY_VAL[bid->first], bid_price = bid->first;
        if (ask == SELL.end()) ask_size = 0, ask_price = 999999;
        else ask_size = SELL_VAL[ask->first], ask_price = ask->first;
        if (bid_price < ask_price) {
            printf("QUOTE %d %d - %d %d\n", bid_size, bid_price, ask_size, (ask_price == 999999 ? 99999 : ask_price));
            return;
        }
        auto sizeb(LIST[*bid->second.begin()].size), sizea(LIST[*ask->second.begin()].size);
        const auto mini(min(sizeb, sizea));
        printf("TRADE %d %d\n", mini, flag ? ask_price : bid_price);

        auto& b(bid->second), &a(ask->second);
        BUY_VAL[bid->first] -= mini;
        if (!b.empty() && (LIST[*b.begin()].size -= mini) == 0)
            b.erase(b.begin());
        if (b.empty()) BUY.erase(bid);
        SELL_VAL[ask->first] -= mini;
        if (!a.empty() && (LIST[*a.begin()].size -= mini) == 0)
            a.erase(a.begin());
        if (a.empty()) SELL.erase(ask);
    }
}

int main()
{
#ifdef _XIENAOBAN_
#define gets(T) gets_s(T, 66666)
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif

    while (scanf("%d", &N) != EOF) {
        if (T++) puts("");
        LIST.clear(), BUY.clear(), SELL.clear(), BUY_VAL.clear(), SELL_VAL.clear();
        LIST.push_back(for_push_back);
        for (int n(1);n <= N;++n) {
            INFO tmp;
            scanf("%s", type);
            if (*type == 'C') {
                int id;
                scanf("%d", &id);
                auto& csl(LIST[id]);
                if (csl.type == 'B') {
                    BUY_VAL[csl.price] -= csl.size;
                    LIST[id].size = 0;
                    auto& now(BUY[csl.price]);
                    now.erase(id);
                    if (now.empty()) BUY.erase(csl.price);
                }
                else {
                    SELL_VAL[csl.price] -= csl.size;
                    LIST[id].size = 0;
                    auto& now(SELL[csl.price]);
                    now.erase(id);
                    if (now.empty()) SELL.erase(csl.price);
                }
            }
            else {
                tmp.type = *type;
                scanf("%d%d", &tmp.size, &tmp.price);
                if (*type == 'B') BUY[tmp.price].insert(n), BUY_VAL[tmp.price] += tmp.size;
                else SELL[tmp.price].insert(n), SELL_VAL[tmp.price] += tmp.size;
            }
            LIST.push_back(std::move(tmp));
            trade(*type == 'B');
        }
    }
    return 0;
}

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分析：书上推荐使用优先队列，然而并没有想出来怎么去用。想了想还是用了map，用它的begin()，效果一样的，还可以灵活差入数据。一开始老是在某组测试数据上出现Runtime error，搞得生无可恋。（从一老司机学到的新技能：当提交OJ出现RE时（WA也行），在执行每组数据计算的代码末尾加一句“while(1);”再提交如果从RE变成TLE了则说明是某些特殊数据没照顾到。毕竟不是所有OJ都有udebug可以用，这招还是不错的）

然后过了一天还是心里放不下，又回来想了想，是题目中一个细节“If there is no active order to sell, then it is assumed that ask size is zero and ask price is 99 999. Note, that zero is not a legal price, but 99 999 is a legal price. Recipient of quote messages distinguishes actual 99 999 ask price from the special case of absent orders to sell by looking at its ask size.”出了问题。于是把不可交易价格改成999999，与交易最大价格加以99999区分，果然似乎没问题了。然而接下来就有新问题，Time limit exceeded。。。

又过了一天，心里还是放不下，于是去网上查了查别人的代码，他们不仅用两个map存买卖信息，还再单独开辟两个map来存放sell与buy的总size（就相当于我上面那两个BUY_VAL 和 SELL_VAL），而当时我只用了两个map（BUY和SELL）存买卖的所有信息，所以每次查询循环总的size时要经历一个比较烦的循环，就是以下这个循环：

//求当前价格总的size函数
int sum(set<int>& now) {
    int re(0);
    for (const auto& r : now) re += LIST[r].size;
    return re;
}

//trade函数片段
auto bid(BUY.begin()), ask(SELL.begin());
if (bid == BUY.end()) bid_size = 0, bid_price = 0;
else bid_size = sum(bid->second), bid_price = bid->first;
if (ask == SELL.end()) ask_size = 0, ask_price = 999999;
else ask_size = sum(ask->second), ask_price = ask->first;

这就是超时的罪魁祸首。改进了下瞬间变成0.330s，还是有些出乎意料。