uva1658

/*
solution:
跟原来模板不同的是要求不能重复访问节点，所以要用到拆点法，将2~v-1的每个结点拆成i和i‘两个点
中间用容量为1费用为0的边连接起来。然后求1～v的流量为2的最小费用即可

note:
拆点法

date:
2016/4/24
*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
#include <cstring>

using namespace std;
const int maxn = 2000 + 10;
const int INF = 1000000000;

int n, m;

struct Edge {
int from, to, cap, flow, cost;
Edge(int u, int v, int c, int f, int w):from(u),to(v),cap(c),flow(f),cost(w) {}
};

struct MCMF {
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn];         // 是否在队列中
int d[maxn];           // Bellman-Ford
int p[maxn];           // 上一条弧
int a[maxn];           // 可改进量

void init(int n) {
this->n = n;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}

void AddEdge(int from, int to, int cap, int cost) {
edges.push_back(Edge(from, to, cap, 0, cost));
edges.push_back(Edge(to, from, 0, 0, -cost));
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}

bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost) {
for(int i = 0; i < n; i++) d[i] = INF;
memset(inq, 0, sizeof(inq));
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;
for(int i = 0; i < G[u].size(); i++) {
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;
if(flow + a[t] > flow_limit) a[t] = flow_limit - flow;
flow += a[t];
cost += d[t] * a[t];
for(int u = t; u != s; u = edges[p[u]].from) {
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
}
return true;
}

// 需要保证初始网络中没有负权圈
int MincostFlow(int s, int t, int flow_limit, int& cost) {
int flow = 0; cost = 0;
while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
return flow;
}

};

MCMF g;

int main()
{
//freopen("input.txt", "r", stdin);
while(scanf("%d%d", &n, &m) == 2 && n) {
g.init(2 * n - 2);
for(int i = 2; i <= n - 1; i++)
g.AddEdge(i - 1, i + n - 2, 1, 0);

int a, b, val;
while(m--) {
scanf("%d%d%d", &a, &b, &val);
if(a != 1 && a != n)
a += n - 2;
else
a--;
b--;

g.AddEdge(a, b, 1, val);
}

int cost = 0;
g.MincostFlow(0, n-1, 2, cost);
printf("%d\n", cost);
}
return 0;
}