codevs 1269 匈牙利游戏
2012年CCC加拿大高中生信息学奥赛
Welcome to the Hungary Games! The streets of Budapest form a twisted network of one-way streets.
欢迎来到匈牙利游戏！布达佩斯（匈牙利首都）的街道形成了一个弯曲的单向网络。
You have been forced to join a race as part of a “Reality TV” show where you race through these streets, starting at the Sz´echenyi thermal bath (s for short) and ending at the Tomb of G¨ ul Baba (t for short).
你被强制要求参加一个赛跑作为一个TV秀的一部分节目，比赛中你需要穿越这些街道，从s开始，到t结束。
Naturally, you want to complete the race as quickly as possible, because you will get more promo- tional contracts the better you perform.
很自然的，你想要尽快的完成比赛，因为你的比赛完成的越好，你就能得到更多的商业促销合同。
However, there is a catch: any person who is smart enough to take a shortest s-t route will be thrown into the P´alv¨olgyi cave system and kept as a national treasure. You would like to avoid this fate, but still be as fast as possible. Write a program that computes a strictly-second-shortest s-t route.
但是，有一个需要了解的是，如果有人过于聪明找到从s到t的最短路线，那么他就被扔到国家*人类保护系统中作为一个国家宝藏收藏起来。你显然要避免这种事情的发生，但是也想越快越好。写一个程序来计算一个从s到t的严格次短路线吧。
Sometimes the strictly-second-shortest route visits some nodes more than once; see Sample Input 2 for an example.
有的时候，严格次短路线可能访问某些节点不止一次。样例2是一个例子。
The first line will have the format N M, where N is the number of nodes in Budapest and M is the number of edges. The nodes are 1,2,...,N; node 1 represents s; node N represents t. Then there are M lines of the form A B L, indicating a one-way street from A to B of length L. You can assume that A != B on these lines, and that the ordered pairs (A,B) are distinct.
第一行包含两个整数N和M，N代表布达佩斯的节点个数，M代表边的个数。节点编号从1到N。1代表出发点s，N代表终点t。接下来的M行每行三个整数A B L，代表有一条从A到B的长度为L的单向同路。你可以认为A不等于B，也不会有重复的(A,B)对。
Output the length of a strictly-second-shortest route from s to t. If there are less than two possible lengths for routes from s to t, output −1.
输出从s到t的严格次短路的长度。如果从s到t的路少于2条，输出-1。
样例输入1：
4 6
1 2 5
1 3 5
2 3 1
2 4 5
3 4 5
1 4 13
样例输入2：
2 2
1 2 1
2 1 1
样例输出1：
11
样例输出2：
3
对于样例1：
There are two shortest routes of length 10 (1 → 2 → 4,1 → 3 → 4) and the strictly-second- shortest route is 1 → 2 → 3 → 4 with length 11.
对于样例2：
The shortest route is 1 → 2 of length 1, and the strictly-second route is 1 → 2 → 1 → 2 of length 3.
/*直接利用SPFA维护一个点到另一个点的最短路和次短路，维护方法如下：
1、如果from的最短路能更新to的最短路，就让更新之前的最短路等于次短路，然后去更新最短路。
2、如果from的最短路不能跟新to的最短路，但是可以更新次短路，就去更新次短路。
3、如果form的最短路不能跟新to的最短路，也不能更新次短路，但是from的次短路可以更新to的次短路，那么就去更新次短路。 */
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
#include<cstdio>
#define N 50010
#define inf (1<<30)-1
bool inque[N]={};
int n,m,a,b,l;
long long dis[N],cdis[N];
int head[N];
struct Edge{
int v,w,last;
}edge[];
int t=;
void add_edge(int u,int v,int w)
{
++t;
edge[t].v=v;
edge[t].w=w;
edge[t].last=head[u];
head[u]=t;
}
void input()
{
cin>>n>>m;
//scanf("%d%d",&n,&m);
for(int i=;i<=m;++i)
{
cin>>a>>b>>l;
//scanf("%d%d%d",&a,&b,&l);
add_edge(a,b,l);
}
for(int i=;i<=n;++i)
dis[i]=cdis[i]=inf;
}
void spfa(int k)
{
queue<int>Q;
Q.push(k);
inque[k]=true;
dis[k]=;//
while(!Q.empty())
{
int x=Q.front();
Q.pop();
inque[x]=false;
for(int l=head[x];l;l=edge[l].last)
{
if(dis[x]+edge[l].w<dis[edge[l].v])
{
cdis[edge[l].v]=dis[edge[l].v];
dis[edge[l].v]=dis[x]+edge[l].w;
if(!inque[edge[l].v])
{
inque[edge[l].v]=true;
Q.push(edge[l].v);
}
}
else if(edge[l].w+dis[x]>dis[edge[l].v]&&dis[x]+edge[l].w<cdis[edge[l].v])//
{
cdis[edge[l].v]=dis[x]+edge[l].w;
if(!inque[edge[l].v])
{
inque[edge[l].v]=true;
Q.push(edge[l].v);
}
}
else if(cdis[edge[l].v]>cdis[x]+edge[l].w)//
{
cdis[edge[l].v]=cdis[x]+edge[l].w;
if(!inque[edge[l].v])
{
inque[edge[l].v]=true;
Q.push(edge[l].v);
}
}
}
}
}
int main()
{
input();
spfa();
if(cdis[n]==inf) cout<<-;
else cout<<cdis[n]<<endl;
return ;
}