luogu P2190 小Z的车厢

题目链接：luogu P2190 小Z的车厢

题目大意：

题解：
一开始认为很简单，就是一个普通的模拟，然后全WA。

// 错误代码
#include <iostream>
using namespace std;
#define io_speed_up ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define N 1000000 + 10

int n, m, up[N], down[N], train, sum;

int main() {
	io_speed_up;
	cin >> n >> m;
	for (int i = 1; i <= m; ++i) {
		int x, y, z;
		cin >> x >> y >> z;
		up[x] += z;
		down[y] += z;
	}
	for (int i = 1; i <= n; ++i) {
		train += up[i] - down[i];
		sum = sum >= train ? sum : train;
	}
	if (sum % 36) {
		sum = sum / 36 + 1;
	} else {
		sum /= 36;
	}
	cout << sum;
	return 0;
}

发现题目里的细节“环形铁轨”，所以存在(x)比(y)大的情况，所以转换一下思想，假如(x)比(y)大，那就相当于在(x)上车，在(n+1)下车，再在(1)上车，在(y)下车。（由于到(n)就结束，所以(n+1)的下车可以忽略）

#include <iostream>
using namespace std;
#define io_speed_up ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define N 1000000 + 10

int n, m, up[N], down[N], train, sum;

int main() {
	io_speed_up;
	cin >> n >> m;
	for (int i = 1; i <= m; ++i) {
		int x, y, z;
		cin >> x >> y >> z;
		up[x] += z;
		down[y] += z;
		if (x > y) {
			up[1] += z;
		}
	}
	for (int i = 1; i <= n; ++i) {
		train += up[i] - down[i];
		sum = sum >= train ? sum : train;
	}
	if (sum % 36) {
		sum = sum / 36 + 1;
	} else {
		sum /= 36;
	}
	cout << sum;
	return 0;
}