poj3255 Roadblocks

/*
  解析可看：
  http://blog.csdn.net/arthurfree/article/details/37884463
  
  这次debug时，发现犯了个隐蔽的小错误，让我找了好久才发现...
  for (int i = 0; i < G[v].size(); i++) 
  我一开始，把这句里的v，手误打成了i，然后检查时还一直没能发现T^T
*/

#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int N, R;
int f, t, c;
const int MAX_N = 1e5 + 10;
const int INF = 1e6;
struct edge
{
	int to, cost;
	edge (int t, int c)
	{
		to = t; cost = c;
	}
};
typedef pair <int, int> P;
vector<edge> G[MAX_N]; //图的邻接表表示

int dist[MAX_N]; // 最短距离
int dist2[MAX_N]; // 次短距离

void solve()
{
	priority_queue<P, vector<P>, greater<P> > que;
	fill (dist, dist + N, INF);
	fill (dist2, dist2 + N, INF);
	dist[0] = 0;
	que.push( P(0,0) );
	
	while (!que.empty())
	{
		P p = que.top(); que.pop();
		int v = p.second, d = p.first; // v为标号，d为出队的边的cost
		if ( dist2[v] < d ) continue;
		for (int i = 0; i < G[v].size(); i++) 
		{
			edge&e = G[v][i];
			int d2 = d + e.cost;
			if (dist[e.to] > d2) 
			{
				swap(dist[e.to], d2); //巧妙，一箭双雕，将判断是否更新次小值时，还要再次用到的d，一并更新，否则要写三变量交换了
				que.push( P(dist[e.to], e.to) );
			}
			if (dist2[e.to] > d2 && d2 > dist[e.to])
			{
				dist2[e.to] = d2;
				que.push( P(dist2[e.to], e.to) );
			}
 		}	
	}
	cout << dist2[N - 1] << endl;
}
int main()
{
	cin >> N >> R;
	for (int i = 0; i < N; i++)
	G[i].clear();
	
	for (int i = 0; i < R; i++)
	{
//		edge now; int from;
//		cin >> from >> now.to >> now.cost;
//		G[from].push_back(now);
		
		cin >> f >> t >> c;
		f--; t--;
		G[f].push_back(edge(t, c));
		G[t].push_back(edge(f, c));
	}
	solve();
	return 0;
}