題目：

程式碼：

#include <iostream>
#include<queue>
#include<vector>
using namespace std;

int n, m;
bool visit[100001];
structdist
{
	int id;
	int len;
};
distd, d2;
queue<dist>q[100001];
distans[100001];
structcmp
{
	bool operator()(dista, distb)
	{
		return a.len > b.len;
	}
};
priority_queue<
 
dist, vector<dist>, cmp>que;

int main()
{
	
	int a, b, c;
	cin >> n >> m;
	for (int i = 1; i <= n; i++)while (!q[i].empty())q[i].pop();
	for (int i = 1; i <= n; i++)
	{
		ans[i].id = i;
		ans[i].len = 1000000;
	}
	ans[1].len = 0;
	while (!que.empty())que.pop();
	for (int i = 1; i <= n; i++)que.push(ans[i]);
	for (int i = 1; i <= n; i++)visit[i] = false;
	while (m--)
	{
		scanf("%d%d%d", &a, &b, &c);
		d.id = b, d.len = c;
		q[a].push(d);
		d.id = a;
		q[b].push(d);
	}
	while (!que.empty())
	{
		d = que.top();
		que.pop();
		if (d.len > ans[d.id].len)continue;
		if (visit[d.id])continue;
		visit[d.id] = true;
		while (!q[d.id].empty())
		{
			d2 = q[d.id].front();
			q[d.id].pop();
			if (visit[d2.id])continue;
			if (d2.len < ans[d.id].len)d2.len = ans[d.id].len;
			if (ans[d2.id].len>d2.len)ans[d2.id].len = d2.len;
			que.push(d2);
		}
	}
	cout << ans[n].len;
	return 0;
}

所以，整個程式的複雜度是O（n+m），n是頂點數，m是邊數