Roadblocks

       POJ - 3255 

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

Input
Line 1: Two space-separated integers:  N and  R 
Lines 2..  R+1: Each line contains three space-separated integers:  A,  B, and  D that describe a road that connects intersections  A and  B and has length  D (1 ≤  D ≤ 5000)
Output
Line 1: The length of the second shortest path between node 1 and node  N
Sample Input 
4 4
1 2 100
2 4 200
2 3 250
3 4 100
Sample Output 
450
Hint
Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

                 这是一道求次短路的模板题~~大体上思路就是做两个数组d,d2,一个是表示从1到s的最短路径~一个是从n到s的最短路径~~然后最后枚举每两个有关系的段点(a,b)~那关于他们的最近的从1到n的距离就是d[a]+d2[b]+ab的长度~~吧每个距离与d[n](也就是整个图的最短路)相比较~~找出大于d[n]的最小的距离就好了；

                  但是令我疑惑的是：：假设到n有三条路，路径分别是 5 5 6，那么次短路是6而不是5（验证过）~~感觉很别扭；

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
using namespace std;
int m, n;
int head[100000];
#define inf 0x3f3f3f3f
struct edge
{
	int to;
	int ne;
	int len;
}ed[500000];
bool vis[100000];
int d[100000];
int d2[100000];
int cnt = 0;
void init()
{
	for (int s = 0; s <= 9999; s++)
	{
		head[s] = -1;

	}
	for (int s = 0; s < 490000; s++)
	{
		ed[s].ne = -1;
	}
	for (int s = 0; s <= n; s++)
	{
		d[s] = inf;
		d2[s] = inf;
	}
	memset(vis, 0, sizeof(vis));
}
void add(int from, int to, int len)
{
	ed[cnt].to = to;
	ed[cnt].len = len;
	ed[cnt].ne = head[from];
	head[from] = cnt++;
}
void spfa(int e,int d[])
{
	d[e] = 0;
	queue<int>q;
	q.push(e);
	vis[e] = 1;
	while (!q.empty())
	{
		int t = q.front();
		q.pop();
		for (int s = head[t]; ~s; s = ed[s].ne)
		{
			if (d[ed[s].to] > d[t] + ed[s].len)
			{
				d[ed[s].to] = d[t] + ed[s].len;
				if (!vis[ed[s].to])
				{
					vis[ed[s].to] = 1;
					q.push(ed[s].to);
				}
			}
		}
		vis[t] = 0;
	}
}
int main()
{
	while (cin >> n >> m)
	{
		cnt = 0;
		init();
		for (int s = 0; s < m; s++)
		{
			int a, b, c;
			cin >> a >> b >> c;
			add(a, b, c);
			add(b, a, c);
		}
		spfa(1,d);
		spfa(n, d2);
		int ans = inf;
		for (int s = 1; s <= n; s++)
		{
			for (int t = head[s]; ~t; t = ed[t].ne)
			{
				int a = ed[t].to;
				int le = ed[t].len;
				int te = d[s] + d2[a] + le;
				if (te > d[n] && ans > te)
				{
					ans = te;
				}
			}
		}
		cout << ans << endl;
	}
	return 0;
}