如何降低时间复杂度啊,没有过的数据全都是TLE
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  • 楼主Rinkan
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  • 发布时间2024/5/20 13:51
  • 上次更新2024/5/20 18:57:12
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如何降低时间复杂度啊,没有过的数据全都是TLE
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Rinkan楼主2024/5/20 13:51
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>

using namespace std;
const int N = 1510, M = 5e4 + 10;
int n, m;
struct Node
{
	int id;
	int time;
	Node* next;
	Node(int _id) :id(_id), next(NULL) {}
}*head[N];
int d[N], q[N];
long long int spend[N];
void add(int a, int b, int time)
{
	auto p = new Node(b);
	p->time = time;
	p->next = head[a];
	head[a] = p;
}
bool topsort()
{
	int hh = 0, tt = -1;
	q[++tt] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (!d[i])
		{
			for (auto p = head[i]; p; p->next) --d[p->id];
		}
	}
	while (hh <= tt)
	{
		int t = q[hh++];
		for (auto p = head[t]; p; p = p->next)
		{
			spend[p->id] = max(spend[p->id], spend[t] + p->time);
			if (--d[p->id] == 0) q[++tt] = p->id;
		}
	}
	return tt == n - 1;
}
int main()
{
	cin >> n >> m;
	//for (int i = 1; i <= n; i++) head[i]->time = 0;
	while (m--)
	{
		int a, b, c;
		cin >> a >> b >> c;
		add(a, b, c);
		d[b]++;
	}
	auto ans = topsort();
	if (ans == false) cout << "-1" << endl;
	else cout << spend[n] << endl;
	return 0;
}
2024/5/20 13:51
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