/*
translation:
给出一张图,求从1~n再从n~1的最短路径是多少?注意同一条路径不能走两次!
solution:
最小费用最大流
这个问题可以转化为求2条从1~n不相交的路径,进而就是直接求从1~n的流量为2的最小费用流。
note:
* 本质上是求从1~n的流量为2的最小费用流
# 不能直接用两次dijkstra来求。(将第一次走过的路删掉再跑第二次)
研究问题模型的时候别被题意牵着走,要看出本质的模型。
*/
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 1000 + 5;
const int INF = 0x3f3f3f3f;
typedef pair P;
struct Edge
{
int to, cap, cost, rev;
Edge(int to_, int cap_, int cost_, int rev_):to(to_),cap(cap_),cost(cost_),rev(rev_){}
};
vector G[maxn];
int V, n, m;
int h[maxn], dist[maxn], prevv[maxn], preve[maxn];
void add_edge(int from, int to, int cap, int cost)
{
G[from].push_back(Edge(to, cap, cost, G[to].size()));
G[to].push_back(Edge(from, 0, -cost, G[from].size()-1));
}
int min_cost_flow(int s, int t, int f)
{
int res = 0;
memset(h, 0, sizeof(h));
while(f > 0) {
priority_queue, greater
> pq;
fill(dist, dist + V, INF);
dist[s] = 0;
pq.push(P(0, s));
while(!pq.empty()) {
P p = pq.top(); pq.pop();
int v = p.second;
if(dist[v] < p.first) continue;
for(int i = 0; i < G[v].size(); i++) {
Edge& e = G[v][i];
if(e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
prevv[e.to] = v;
preve[e.to] = i;
pq.push(P(dist[e.to], e.to));
}
}
}
if(dist[t] == INF) return -1;
for(int v = 0; v < V; v++) h[v] += dist[v];
int d = f;
for(int v = t; v != s; v = prevv[v]) {
d = min(d, G[prevv[v]][preve[v]].cap);
}
f -= d;
res += d * h[t];
for(int v = t; v != s; v = prevv[v]) {
Edge& e = G[prevv[v]][preve[v]];
e.cap -= d;
G[v][e.rev].cap += d;
}
}
return res;
}
int main()
{
//freopen("in.txt", "r", stdin);
while(~scanf("%d%d", &n, &m)) {
for(int i = 0; i < maxn; i++) G[i].clear();
int u, v, c;
for(int i = 0; i < m; i++) {
scanf("%d%d%d", &u, &v, &c);
add_edge(u - 1, v - 1, 1, c);
add_edge(v - 1, u - 1, 1, c);
}
int s = 0, t = n - 1; V = n;
printf("%d\n", min_cost_flow(s, t, 2));
}
return 0;
}