出题人题解：(O(m^2))

class Solution {
public:
    bool carPooling(vector<vector<int>>& trips, int capacity) {
		int all = 0;//判断会不会超过最大载客量
		int destination=trips[0][2];
		for (int i = 0; i < trips.size(); i++)//找到最远的地点
		{
			all += trips[i][0];
			if (trips[i][2] > destination)
				destination = trips[i][2];
		}

		if (all <= capacity)
			return true;



		all = 0;
		for (int i = 0; i < destination; i++)//不断循环测试每前进一站会发生什么
		{
			for (int k = 0; k < trips.size(); k++)//是否有人下车上车
			{
				if (trips[k][2] == i)//下车
				{
					all -= trips[k][0];
				}

				if (trips[k][1] == i)//上车
				{
					all += trips[k][0];
				}

			}
			if (all > capacity)
				return false;

		}

		return true;
    }
};

更优解：O(m)

#include <bits/stdc++.h>
using namespace std;
#define sc(x) scanf("%lld", &x)
typedef long long ll;
const ll mn = 1010;
ll n, m, s[mn], in[mn], out[mn], a, b, c, cnt, tp;
signed main()
{
    sc(m), sc(n);
    while (m--)
    {
        sc(a), sc(b), sc(c);
        in[b] += a, out[c] += a;
        tp = max({tp, b, c});
    }
    for (ll i = 1; i <= tp; ++i)
    {
        cnt = cnt - out[i] + in[i];
        if (cnt > n)
        {
            printf("0");
            return 0;
        }
    }
    printf("1");
    return 0;
}