UVa 1658 Admiral (最小费用流)

题意：给定一个图，求1-n的两条不相交的路线，并且权值和最小。

析：最小费用流，把每个结点都拆成两个点，中间连一条容量为1的边，然后一个作为入点，另一个是出点。最后跑两次最小费用流就行了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 2000 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
  int from, to, cap, flow;
  LL cost;
};
 
struct MCMF{
  int n, m;
  vector<Edge> edges;
  vector<int> G[maxn];
  int inq[maxn];
  LL d[maxn];
  int p[maxn];
  int a[maxn];
 
  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, LL 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, LL &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;
    flow += a[t];
    cost += d[t] * a[t];
    int u = t;
    while(u != s){
      edges[p[u]].flow += a[t];
      edges[p[u]^1].flow -= a[t];
      u = edges[p[u]].from;
    }
    return true;
  }
 
  LL minCost(int s, int t){
    int flow = 0;
    LL cost = 0;
    while(BellmanFord(s, t, flow, cost));
    return cost;
  }
};
MCMF mcmf;
 
int main(){
  while(scanf("%d %d", &n, &m) == 2){
    int u, v, val;
    mcmf.init(n+n+1);
    for(int i = 0; i < m; ++i){
      scanf("%d %d %d", &u, &v, &val);
      mcmf.addEdge(u+n, v, 1, val);
    }
    for(int i = 2; i < n; ++i)
      mcmf.addEdge(i, i+n, 1, 0);
    mcmf.addEdge(1, 1+n, 2, 0);
    mcmf.addEdge(n, n+n, 2, 0);
    printf("%lld\n", mcmf.minCost(1, n+n) + mcmf.minCost(1, n+n));
  }
  return 0;
}