说一下几种做法：

1. y总的单调队列做法，45m可以看成滑动窗口，用单调队列维护。
2. 枚举地铁，尝试优惠公交
3. 枚举公交，尝试使用地铁优惠

我选择的是枚举地铁，时间判断。

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int n;
struct Node
{
    int c, price, t;
}alls[N];
int main()
{
    long long int sum = 0;
    cin >> n; 
    for(int i = 0; i < n; i ++) 
    {
        int op, op2, op3;
        scanf("%d%d%d", &op, &op2, &op3);
        alls[i].c = op, alls[i].price = op2, alls[i].t = op3;
        if(alls[i].c == 0) sum += alls[i].price;
    }
    for(int i = 0; i < n; i ++)
        if(alls[i].c == 0)
            for(int j = i + 1; j < n; j ++)
            {
                if(alls[j].t - alls[i].t <= 45 && alls[i].price >= alls[j].price && alls[j].c == 1 && alls[j].price != 0)
                {
                    alls[i].price = 0, alls[j].price = 0;
                    break;
                }
                if(alls[j].t - alls[i].t > 45) break;
            }
    for(int i = 0; i < n; i ++)
        if(alls[i].c == 1)
            sum += alls[i].price;
    printf("%lld", sum);
}

顺手给一个调试器：

/*
if( i == 2 && j == 3) 
{
    cout << alls[i].c << ' ' << alls[i].price << ' ' << alls[i].t << ' ' << alls[j].c << ' ' << alls[j].price << ' ' << alls[j].t << endl;
    if(alls[j].t - alls[i].t <= 45) cout << "1Yes ";
    if(alls[i].price >= alls[j].price) cout << "2Yes ";
    if(alls[j].c == 1) cout << "3Yes ";
    if(alls[j].price != 0) cout << "4Yes ";
    cout << endl;
}
*/

update: 单调队列解法AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>

using namespace std;

const int N = 100010;

int n;

struct Node
{
    int c, price, t;
}alls[N];

pair<int, int> q[N];
int hh = 0, tt = -1;
bool st[N]; 

int main()
{
    cin >> n;

    for(int i = 0; i < n; i ++)
    {
        cin >> alls[i].c >> alls[i].price >> alls[i].t;
    }

    long long int res = 0;

    for(int i = 0; i < n; i ++)
    {
        if(alls[i].c == 0) 
        {
            res += alls[i].price;
            q[ ++ tt ] = {alls[i].price, alls[i].t};
        }
        else
        {
            bool flag = false;
            while(hh <= tt && alls[i].t - q[hh].second > 45) hh ++;
            for(int j = hh; j <= tt; j ++)
                if(q[j].first >= alls[i].price && st[j] == false)
                {
                    st[j] = true;
                    flag = true;
                    break;
                }
            if(flag == false) res += alls[i].price;
        }
    }

    cout << res << endl;

    return 0;
}