2019 10th Lanqiao Cup Provincial Competition C++ University Group A Question G Takeaway Store Priority [Simulation]

Portal: AcWing 1241. Takeaway priority / Lanqiao Cup 2019 Tenth Real Question

Ideas

Summarize orders at the same time, and then traverse each time.

Only consider the moment when there is an order, for moment $i$ , first summarize the number of orders in the same store, and then traverse all orders, each order contains store number information (assuming the store number is $i d$ ), maintain store $i d The$ time when the order last appeared (assuming it is $l a s t [i d]$ ), then the shop $i d$ 之前有 $i - l a s t [i d] -$ There is no order at the time of $1$ , at this time, the priority needs to be subtracted from the time when there is no order. If the priority is <=3, the store $i d$ move out of the cache queue; then the priority plusPriority brought by $n$ $u$ $m$ orders $2 * n u m$ , if the priority is >5, enter the buffer queue.

Finally, you need to traverse all Shops with no orders at time $T, the$ time they last order appears to $The$ period of time $T$ is the reduced value of its priority, and at the same time whether the update is removed from the cache queue.

$T$ moments, $m$ orders, $n$ stores, the time complexity is about $O (T + m + n)$ because for less than $At$ the moment of $T$ , the priority of all stores is not updated, but only the priority of the stores with orders is updated, only in the last $It is$ sufficient to update the priority of all stores once at $T.$

AC code

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int n,m,T,last[N],sum[N];
bool f[N]; // 是否在缓存队列中
vector<int>ans[N];
unordered_map<int,int>vis;
int main()
{
    
    
    ios::sync_with_stdio(false);
    cin>>n>>m>>T;
    int t,id;
    for(int i=1;i<=m;i++)
    {
    
    
        cin>>t>>id;
        ans[t].push_back(id); // t时刻编号为id的店有订单
    }
    memset(last,-1,sizeof(last));
    for(int i=1;i<=T;i++) // 遍历时间
    {
    
    
        if(ans[i].size()) // i时刻有订单
        {
    
    
            int sz=ans[i].size();
            vis.clear();
            for(int j=0;j<sz;j++)
            {
    
    
                int id=ans[i][j];
                vis[id]++; // 汇总相同店的订单数
            }
            for(auto it:vis)
            {
    
    
                int id=it.first; // 店id
                int num=it.second; // 订单数num
                if(last[id]==-1) // 店id第1次出现
                {
    
    
                    sum[id]=num*2;
                    last[id]=i;
                    if(sum[id]>5)f[id]=1; // 放入缓存队列
                }
                else
                {
    
    
                    int d=i-last[id]-1;
                    sum[id]=max(0,sum[id]-d);
                    if(sum[id]<=3)f[id]=0; // 移出缓存队列
                    sum[id]+=num*2;
                    if(sum[id]>5)f[id]=1; // 放入缓存队列
                    last[id]=i;
                }
            }
        }
        if(i==T) // 处理T时刻无订单的店
        {
    
    
            for(int j=1;j<=n;j++) 
            {
    
    
                if(last[j]!=T) // 店j的最后一次订单出现在T时刻之前
                {
    
    
                    int d=T-last[j]; 
                    sum[j]=max(0,sum[j]-d);
                   if(sum[j]<=3)f[j]=0; // 移出缓存队列
                }
            }
        }
    }
    int cnt=0;
    for(int i=1;i<=n;i++)
        if(f[i])cnt++;
    printf("%d\n",cnt);
    return 0;
}
/*
2 6 6
1 1
5 2
3 1
6 2
2 1
6 2
ans:1
2 6 8
1 1
5 2
3 1
6 2
2 1
6 2
ans:1
2 6 9
1 1
5 2
3 1
6 2
2 1
6 2
ans:0
1 3 5
1 1
2 1
3 1
ans:1
1 3 6
1 1
2 1
3 1
ans:0
*/