CCF 201412-3 集合竞价

CCF 201412-3 集合竞价 传送门

只得了80分, 硬是找不出bug在哪里.

思路是:
1. 对命令进行处理, 改取消的取消, 可以stringstream方便读取字符串中的string double int.
2. 将两种命令(sell和buy)分别按价格排序.
3. 同时向右查询, 以buy为循环外围, 对每个可更新答案的情况进行更新. 答案一定是buy的某个价格, 对每一个价格, 找它的尽可能成交量多的情况.

我估计我是步骤三有个bug

80分Code

#include <algorithm>
#include <iostream>
#include <sstream>
#include <vector>
using namespace std;

struct Node {
    double money;
    int num;
    Node (double m, int n): money(m), num(n) {}
};

bool cmp(Node x, Node y)
{
    return x.money < y.money;
}

string str[5005];
int sum_buy[5005], sum_sell[5005];

int main()
{
    int t = 0;
    while (getline(cin, str[++t])) {
        if (str[t][0] == 'c') {
            int in = 7, line = 0;
            for (int i = in; i < str[t].size(); ++i) {
                line = line*10 + int(str[t][i] - '0');
            }
            str[line] = "";
            str[t] = "";
        }
    }
    vector<Node> sell, buy;
    for (int i = 1; i <= t; ++i) {
        stringstream sin(str[i]);
        string tmp;
        double money;
        int num;
        sin >> tmp >> money >> num;
        if (str[i][0] == 's') sell.push_back(Node(money, num));
        else if (str[i][0] == 'b') buy.push_back(Node(money, num));
    }
    sort(buy.begin(), buy.end(), cmp);
    sort(sell.begin(), sell.end(), cmp);
    sum_buy[buy.size()] = 0;
    for (int i = buy.size() - 1; i >= 0; --i)
        sum_buy[i] = sum_buy[i + 1] + buy[i].num;
    sum_sell[0] = sell[0].num;
    for (int i = 1; i < sell.size(); ++i)
        sum_sell[i] = sum_sell[i - 1] + sell[i].num;
    double ans_mny = 0;
    long long ans_cnt = 0;
    for (int i = 0, j = 0; i < buy.size(); ++i) {
        double bmo = buy[i].money;
        while (sell[j].money <= bmo) {
            if (j == sell.size() - 1 || sell[j+1].money > bmo) {
                if (ans_cnt <= min(sum_buy[i], sum_sell[j])) {
                    ans_cnt = min(sum_buy[i], sum_sell[j]);
                    ans_mny = bmo;
                }
                break;
            }
            j++;
        }
    }
    printf("%.2lf %d", ans_mny, ans_cnt);
}