UVa 1658 - Admiral（最小费用最大流 + 拆点）

链接：

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4533

题意：

给出一个v（3≤v≤1000）个点e（3≤e≤10000）条边的有向加权图，
求1～v的两条不相交（除了起点和终点外没有公共点）的路径，使得权和最小。

分析：

把2到v-1的每个结点i拆成i和i'两个结点，中间连一条容量为1，费用为0的边，然后求1到v的流量为2的最小费用流即可。
本题的拆点法是解决结点容量的通用方法。

代码：

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <vector>
 using namespace std;

 /// 结点下标从0开始，注意maxn
 struct MCMF {
     static const int maxn =  *  + ;
     static const int INF = 0x3f3f3f3f;
     struct Edge {
         int from, to, cap, flow, cost;
     };

     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;
     }
     // 需要保证初始网络中没有负权圈
     int MincostMaxflow(int s, int t, int flow_limit) {
         int flow = , cost = ;
         //while(BellmanFord(s, t, flow, cost));
         while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
         return cost;
     }
 };

 MCMF mc;

 int main() {
     int v, e;
     while(~scanf("%d%d", &v, &e)) {
         mc.init(v*-);
         for(int i = ; i < v-; i++) mc.AddEdge(i, i+v-, , );
         for(int a, b, c, i = ; i < e; i++) {
             scanf("%d%d%d", &a, &b, &c);
             a = (a ==  || a == v) ? a- : a+v-;
             b--;
             mc.AddEdge(a, b, , c);
         }
         printf("%d\n", mc.MincostMaxflow(, v-, ));
     }
     return ;
 }

最小费用最大流模板：

 /// 结点下标从0开始，注意maxn
 struct MCMF {
     static const int maxn = 1e3 + ;
     static const int INF = 0x3f3f3f3f;
     struct Edge {
         int from, to, cap, flow, cost;
     };

     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, 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;
     }
     // 需要保证初始网络中没有负权圈
     int MincostMaxflow(int s, int t) {
         int flow = , cost = ;
         while(BellmanFord(s, t, flow, cost));
         //while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
         return cost;
     }
 };