第三章 唯一剩下的 —— 外卖店优先级

//结构体排序

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>

using namespace std;
const int N = 1e5 + 10;

int flag[N];  //记录第id个商店的优先级
int tap[N];
int output[N];

struct node
{
    int time;
    int id;
}nodes[N];

bool cmp(node a, node b)
{
    if (a.id != b.id)return a.id < b.id;
    if (a.time != b.time)return a.time < b.time;
    return a.time < b.time;
}

int main()
{
    int n, m, t;
    cin >> n >> m >> t;
    //n个外卖店 m个订单 t时间内
    for (int i = 0; i < m; i++)
        cin >> nodes[i].time >> nodes[i].id;
    sort(nodes, nodes + m, cmp);
//    for (int i = 0; i < m; i++)cout << nodes[i].time << ' ' << nodes[i].id << endl;
    int cnt = 0;
    //计算每一个商店有几个订单
    for (int i = 0; i < m; i++)
    {
        if (nodes[i].id == nodes[i + 1].id)tap[cnt]++;
        else tap[cnt]++, cnt++;
    }

    int k = 0;  //指向有效订单从0 指向m
    for (int i = 0, k = 0; i < n; i++)  //讨论每一个商店
    {
        //每个商店有tap[i]个订单
        for (int j = nodes[k].time, tmp = 0; j <= t; j++)  //一共遍历的时间
        {
      //      cout << flag[i] << ' ' << j << endl;
            if (flag[i] > 5)output[i] = 1;
            if (flag[i] <= 3)output[i] = 0;
            int yy = 0;
            while((j == nodes[k].time) && (tmp < tap[i])) flag[i] += 2, k++,tmp++,yy = 1;
            if (flag[i] > 5)output[i] = 1;
            if (flag[i] <= 3)output[i] = 0;
            if (yy == 1)continue;
            if (tmp >= tap[i]) flag[i] -= 1;
            else if (j != nodes[k].time) flag[i] -= 1;
            if (flag[i] <= 0) flag[i] = 0;
            if (flag[i] > 5)output[i] = 1;
            if (flag[i] <= 3)output[i] = 0;

        }
    }
    int ans = 0;
    for (int i = 0; i < n; i++)
    {
        cout << flag[i] << endl;
        if ( flag[i] >= 4 && output[i] ==1)ans++;  //上过>5 并且现在不能小于
    }
    cout << endl << ans << endl;
    return 0;
}

我一开始的做法：按照他的要求，踢皮球一般看一个要求写一些。没有全局观，导致TEL。