UVa 1658 (拆点法 最小费用流) Admiral

题意：

给出一个有向带权图，求从起点到终点的两条不相交路径使得权值和最小。

分析：

第一次听到“拆点法”这个名词。

把除起点和终点以外的点拆成两个点i和i'，然后在这两点之间连一条容量为1，费用为0的边。这样就保证了每个点最多经过一次。

其他有向边的容量也是1

然后求从起点到终点的流量为2(这样就保证了是两条路径)的最小费用流。

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn =  + ;
 const int INF = ;

 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 inq[maxn]; //是否在队列中
     int d[maxn];    //Bellman-Ford
     int p[maxn];    //上一条弧
     int a[maxn];    //可改进量

     void Init(int n)
     {
         this->n = n;
         for(int i = ; 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, , cost));
         edges.push_back(Edge(to, from, , , -cost));
         m = edges.size();
         G[from].push_back(m-);
         G[to].push_back(m-);
     }

     bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost)
     {
         for(int i = ; i < n; ++i) d[i] = INF;
         memset(inq, , sizeof(inq));
         d[s] = ; inq[s] = ; p[s] = ; a[s] = INF;

         queue<int> Q;
         Q.push(s);
         while(!Q.empty())
         {
             int u = Q.front(); Q.pop();
             inq[u] = ;
             for(int i = ; 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] = ; }
                 }
             }
         }
         if(d[t] == INF) return false;
         if(flow + a[t] > flow_limit) a[t] = flow_limit - flow;
         flow += a[t];
         cost += d[t] * a[t];
         for(int u = t; u != s; u = edges[p[u]].from)
         {
             edges[p[u]].flow += a[t];
             edges[p[u]^].flow -= a[t];
         }
         return true;
     }

     int MincostMaxflow(int s, int t, int flow_limit, int& cost)
     {
         int flow = ; cost = ;
         while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
         return flow;
     }
 }g;

 int main()
 {
     //freopen("in.txt", "r", stdin);

     int n, m;
     while(scanf("%d%d", &n, &m) ==  && n)
     {
         g.Init(n*-);
         //2~n-1 i和i'的编号分别为1~n-2  n~2n-3
         for(int i = ; i <= n-; ++i) g.AddEdge(i-, n-+i, , );
         for(int i = ; i < m; ++i)
         {   //连接a'->b
             int a, b, c;
             scanf("%d%d%d", &a, &b, &c);
             if(a !=  && a != n) a += n-; else a--;
             b--;
             g.AddEdge(a, b, , c);
         }
         int cost;
         g.MincostMaxflow(, n-, , cost);
         printf("%d\n", cost);
     }

     return ;
 }

代码君