When FJ’s friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000.
To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again.
He wants his tour to be as short as possible, however he doesn’t want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.
Input
* Line 1: Two space-separated integers: N and M.
- Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path’s length.
Output
A single line containing the length of the shortest tour.
Sample Input
4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2
Sample Output
6
题意： 从1走到n，从n走到1，要求不走重边情况下的最小费用，因为不能走重边所以每个边的容量为1，费用为该路的长度，增加一个源点，汇点，因为一来回，所以1到源点，n到汇点的边，容量为2，费用为0，直接用紫书上的最小费用最大流模板做。
AC的C++程序如下：
#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 1010;
const int inf = 0x3f3f3f3f;
typedef long long ll;
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 visit[maxn]; //是否在队列中
int dist[maxn];
int p[maxn]; //上一条弧
int a[maxn]; //可改进的量
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,int &cost)
{
for (int i = 0; i <= n; i++) dist[i] = inf;
memset(visit, 0, sizeof(visit));
dist[s] = 0; visit[s] = 1; p[s] = 0; a[s] = inf;
queue<int> q;
q.push(s);
while (!q.empty())
{
int u = q.front(); q.pop();
visit[u] = 0;
for (int i = 0; i < (int)g[u].size(); i++)
{
edge e = edges[g[u][i]];
if (e.cap > e.flow&&dist[e.to] > dist[u] + e.cost)
{
dist[e.to] = dist[u] + e.cost;
p[e.to] = g[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if (!visit[e.to]) q.push(e.to), visit[e.to] = 1;
}
}
}
if (dist[t] == inf) return false;
flow += a[t];
cost += dist[t] * a[t];
for (int i = t; i != s; i = edges[p[i]].from)
{
edges[p[i]].flow += a[t];
edges[p[i] ^ 1].flow -= a[t];
}
return true;
}
int mincostmaxflow(int s, int t, int &cost)
{
int flow = 0; cost = 0;
while (bellmanford(s, t, flow, cost));
return flow;
}
};
int main()
{
int n, m;
cin >> n >> m;
mcmf mc;
mc.init(n + 1);
int src, dest, cost;
for (int i = 1; i <= m; i++)
{
cin >> src >> dest >> cost;
mc.addedge(src, dest, 1, cost);
mc.addedge(dest, src, 1, cost);
}
mc.addedge(0, 1, 2, 0);
mc.addedge(n, n + 1, 2, 0);
int c = 0;
mc.mincostmaxflow(0, n+1, c);
cout << c << endl;
return 0;
}