poj 2135 Farm Tour(最小费用流)

思路：

求往返不能经过同一条道路两次，参观路线最小的最小值.可以转话为边的流量为1，总流量为2的最小费用流

约束:

1<= N <= 1000

1<= M <= 10000

1<= ai, bi <= N

1 <= ci <= 35000

/***********************************************
 * Author: fisty
 * Created Time: 2015-08-17 16:09:15
 * File Name   : poj2135.cpp
 *********************************************** */
#include <iostream>
#include <cstring>
#include <deque>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <algorithm>
using namespace std;
#define Debug(x) cout << #x << " " << x <<endl
#define Memset(x, a) memset(x, a, sizeof(x))
const int INF = 0x3f3f3f3f;
typedef long long LL;
typedef pair<int, int> P;
#define FOR(i, a, b) for(int i = a;i < b; i++)
#define lson l, m, k<<1
#define rson m+1, r, k<<1|1
const int MAXN = 10010;
const int MAXM = 100010;
struct Edge
{
    int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s,int t)
{
    queue<int>q;
    for(int i = 0;i < N;i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1;i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow &&
                    dis[v] > dis[u] + edge[i].cost )
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t] == -1)return false;
    else return true;
}
//返回的是最大流， cost存的是最小费用
int minCostMaxflow(int s,int t)
{
    int flow = 0;
    int cost = 0;
    while(spfa(s,t))
    {
        int Min = INF;
        for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
        {
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return cost;
}
int n, M;
int a[MAXM], b[MAXM], c[MAXM];
void solve(){
    int s = n, t = n + 1;
    init(n + 2);
    for(int i = 0;i < M; i++){
        addedge(a[i]-1, b[i]-1, 1, c[i]);
        addedge(b[i]-1, a[i]-1, 1, c[i]);
    }
    addedge(s, 0, 2, 0);
    addedge(n-1, t, 2, 0);
    printf("%d\n", minCostMaxflow(s, t));
}
int main() {
    //freopen("in.cpp", "r", stdin);
    //cin.tie(0);
    //ios::sync_with_stdio(false);
    while(~scanf("%d%d", &n, &M)){  
        for(int i = 0;i < M; i++){    
            scanf("%d%d%d", &a[i], &b[i], &c[i]);
        }
        solve();
    }
    return 0;
}