n个点的无向带权图，求1->n的最短往返路径，不走重复边。
这里涉及到一个知识点：求无向图上s->t的最短路，其实就是费用流。
而求1->n最短往返路径呢？增加源点s，由s到1加弧，容量为2（往返两次），费用为0；而对于原图中的边<u, v>，分别由u到v，由v到u增加容量为1（往返不能走重边），费用为边权的弧。然后跑费用流得到的最小费用便是答案。如果最后求得的最大流小于2，则说明无解。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<fstream>
#include<sstream>
#include<bitset>
#include<vector>
#include<string>
#include<cstdio>
#include<cmath>
#include<stack>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
using namespace std; const int maxn = 111;
const int INF = 1e9;
int n, m, s, t, d[maxn], p[maxn], a[maxn], inq[maxn];
int flow, cost;
struct Edge
{
int from, to, cap, flow, cost;
};
vector<Edge> edges;
vector<int> G[maxn]; inline void init()
{
flow = cost = s = 0, t = n;
REP(i, t+1) G[i].clear(); edges.clear();
} void add(int from, int to, int cap, int cost)
{
edges.PB((Edge){from, to, cap, 0, cost});
edges.PB((Edge){to, from, 0, 0, -cost});
int nc = edges.size();
G[from].PB(nc-2);
G[to].PB(nc-1);
} bool spfa(int& flow, int& cost)
{
REP(i, t+1) d[i] = INF;
CLR(inq, 0);
d[s] = 0, inq[s] = 1, p[s] = 0, a[s] = INF;
queue<int> q; q.push(s);
while(!q.empty())
{
int u = q.front(); q.pop();
inq[u] = 0;
int nc = G[u].size();
REP(i, nc)
{
Edge& e = edges[G[u][i]];
if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if(!inq[e.to]) q.push(e.to), inq[e.to] = 1;
}
}
}
if(d[t] == INF) return false;
flow += a[t], cost += d[t] * a[t];
int u = t;
while(u != s)
{
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
u = edges[p[u]].from;
}
return true;
} int main()
{
while(scanf("%d", &n), n)
{
scanf("%d", &m);
init();
int a, b, c;
add(s, 1, 2, 0);
while(m--)
{
scanf("%d%d%d", &a, &b, &c);
add(a, b, 1, c);
add(b, a, 1, c);
}
while(spfa(flow, cost));
if(flow < 2) puts("Back to jail");
else printf("%d\n", cost);
}
return 0;
}