POJ2135 Farm Tour

嘟嘟嘟


费用流入门题。
其实我也不知道为啥是费用流，不过因为学费用流的时候推这题了我才能想到。


因为每一条路只能走一次，所以容量设为1，路径长度作为费用。
然后从源点向1号节点连一条容量为2，费用为0的边；从(n)号节点向汇点连一条容量为2，费用为0的边。
跑最小费用流即可。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e3 + 5;
const int maxm = 1e4 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, s, t;
struct Edge
{
  int nxt, from, to, cap, c;
}e[maxm << 2];
int head[maxn], ecnt = -1;
void addEdge(int x, int y, int w, int f)
{
  e[++ecnt] = (Edge){head[x], x, y, w, f};
  head[x] = ecnt;
  e[++ecnt] = (Edge){head[y], y, x, 0, -f};
  head[y] = ecnt;
}

queue<int> q;
int dis[maxn], flow[maxn], pre[maxn];
bool in[maxn];
bool spfa(int s, int t)
{
  Mem(dis, 0x3f); Mem(in, 0);
  dis[s] = 0; in[s] = 1; flow[s] = INF;
  q.push(s);
  while(!q.empty())
    {
      int now = q.front(); q.pop(); in[now] = 0;
      for(int i = head[now], v; i != -1; i = e[i].nxt)
	{
	  v = e[i].to;
	  if(e[i].cap > 0 && dis[now] + e[i].c < dis[v])
	    {
	      dis[v] = dis[now] + e[i].c;
	      flow[v] = min(flow[now], e[i].cap);
	      pre[v] = i;
	      if(!in[v]) in[v] = 1, q.push(v);
	    }
	}
    }
  return dis[t] != INF;
}
ll maxFlow = 0, minCost = 0;
void update(int s, int t)
{
  int x = t;
  while(x != s)
    {
      int i = pre[x];
      e[i].cap -= flow[t];
      e[i ^ 1].cap += flow[t];
      x = e[i].from;
    }
  maxFlow += flow[t];
  minCost += (ll)flow[t] * dis[t];
}

void MCMF(int s, int t)
{
  while(spfa(s, t)) update(s, t);
}

int main()
{
  Mem(head, -1);
  n = read(); m = read(); s = 0; t = n + 1;
  for(int i = 1; i <= m; ++i)
    {
      int x = read(), y = read(), f = read();
      addEdge(x, y, 1, f); addEdge(y, x, 1, f);
    }
  addEdge(s, 1, 2, 0); addEdge(n, t, 2, 0);
  MCMF(s, t);
  write(minCost), enter;
  return 0;
}