思想：和求最短路上的点套路一样，spfa先正着求一次，再反着求一次最短路，然后枚举每条边<i,j>找dist_zheng[i] + len<i,j> + dist_fan[j]的第二小值即可！注意不能用邻接矩阵，那样会MLE，应该用邻接表
/*
poj 3255
3808K 266MS
*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#define MAXN 200005
#define MAX_INT 2147483647
using namespace std;
int last[5005], dist_1[5005], dist_2[5005], n, m, gra[5005][5005];
bool mark[MAXN];
struct node
{
int u;
int v;
int w;
int next;
node()
{
u = v = w = next = 0;
}
}edge[MAXN];
void spfa( int dist[5005], int s )
{
queue<int>myQueue;
dist[s] = 0;
memset(mark, false, sizeof(mark));
mark[s] = true;
myQueue.push(s);
while( !myQueue.empty() )
{
int x = myQueue.front();
myQueue.pop();
mark[x] = false;
int t = last[x];
while( t )
{
if( dist[ edge[t].v ] > dist[x] + edge[t].w )
{
dist[ edge[t].v ] = dist[x] + edge[t].w;
if( !mark[ edge[t].v ] )
myQueue.push( edge[t].v );
}
t = edge[t].next;
}
}
}
int main()
{
cin >> n >> m;
for(int i = 1;i <= m;i ++)
{
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
edge[i].u = edge[i + m].v = a;
edge[i].v = edge[i + m].u = b;
edge[i].w = edge[i + m].w = c;
edge[i].next = last[a];
last[a] = i;
edge[i + m].next = last[b];
last[b] = i + m;
}
memset( dist_1, 1, sizeof(dist_1) );
spfa( dist_1, 1 );
memset( dist_2, 1, sizeof(dist_2) );
spfa( dist_2, n );
int ans = MAX_INT, tmp = MAX_INT;
for(int i = 1;i <= n;i ++)
{
int t = last[i];
while( t )
{
if( dist_1[i] + dist_2[ edge[t].v ] + edge[t].w < tmp )
{
ans = tmp;
tmp = dist_1[i] + dist_2[ edge[t].v ] + edge[t].w;
}
else if( dist_1[i] + dist_2[ edge[t].v ] + edge[t].w < ans
&& dist_1[i] + dist_2[ edge[t].v ] + edge[t].w != tmp )
ans = dist_1[i] + dist_2[ edge[t].v ] + edge[t].w;
t = edge[t].next;
}
}
cout << ans << endl;
return 0;
}