【POJ 2135】Farm Tour

1.​​题目链接​​。题目大意：给定一个n个点m条边的无向图，每条边都有一个长度，求从1->n->1走过的长度最小值并且每条边只能走一次。

2.由于边是无向的，其实这个题目就是在问从1到n两条不相交的路径的长度之和最小值。考虑把长度作为费用，就转化成了一个每条边的容量为1的费用里问题，建立源点s，到1的流量为2，费用为0，汇点t。到n的流量为2，费用为0，所以就成了s，到t的MCMF。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<fstream>
#include<math.h>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
const int inf = 1 << 29;
const int MAXN = 1010;
const int MAXM = 40010;
struct Node
{
  int s;
  int to;
  int next;
  int capacity;
  int value;
};

int n, m;
int index;
Node node[MAXM];
int head[MAXN];
int pre[MAXN];
int dis[MAXN];
bool vis[MAXN];

void init()
{
  index = 0;
  memset(head, -1, sizeof(head));
  memset(node, 0, sizeof(node));
}
void addedge(int a, int b, int v, int c)
{
  node[index].to = b;
  node[index].s = a;
  node[index].value = v;
  node[index].capacity = c;
  node[index].next = head[a];
  head[a] = index++;

  node[index].to = a;
  node[index].s = b;
  node[index].value = -v;
  node[index].capacity = 0;
  node[index].next = head[b];
  head[b] = index++;
}
bool spfa(int s, int t, int nnum)
{
  memset(vis, 0, sizeof(vis));
  memset(pre, -1, sizeof(pre));
  for (int i = 0; i <= nnum; i++)
  {
    dis[i] = inf;
  }
  queue<int> que;
  que.push(s);
  dis[s] = 0;
  vis[s] = true;
  while (!que.empty())
  {
    int temp = que.front();
    que.pop();
    vis[temp] = false;
    for (int i = head[temp]; i != -1; i = node[i].next)
    {
      if (node[i].capacity)
      {
        int ne = node[i].to;
        if (dis[temp] + node[i].value < dis[ne])
        {
          dis[ne] = dis[temp] + node[i].value;
          pre[ne] = i;
          if (!vis[ne])
          {
            vis[ne] = true;
            que.push(ne);
          }
        }
      }
    }
  }
  if (dis[t] == inf)
    return false;
  return true;
}
int getMincost(int s, int t, int nnum)
{
  int ans_flow = 0;
  int ans_cost = 0;
  int temp, minc;
  while (spfa(s, t, nnum))
  {
    temp = t;
    minc = inf;
    while (pre[temp] != -1)
    {
      minc = min(node[pre[temp]].capacity, minc);
      temp = node[pre[temp]].s;
    }
    temp = t;
    while (pre[temp] != -1)
    {
      node[pre[temp]].capacity -= minc;
      int ss = pre[temp] ^ 1;
      node[ss].capacity += minc;
      temp = node[pre[temp]].s;
    }
    ans_cost += dis[t] * minc;
  }
  return ans_cost;
}
int main()
{
  int a, b, v;
  int s, t, result;
  while (cin >> n >> m)
  {
    init();
    for (int i = 0; i < m; i++)
    {
      cin >> a >> b >> v;
      //a->b的费用是v,流量是0
      addedge(a, b, v, 1);
      addedge(b, a, v, 1);
    }
    s = n + 1;
    t = s + 1;
    addedge(s, 1, 0, 2);
    addedge(n, t, 0, 2);
    result = getMincost(s, t, t);
    cout << result << endl;
  }
  return 0;
}