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。

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 const int maxn = 2000 + 10;
  6 const int INF = 1000000000;
  7 
  8 struct Edge
  9 {
 10     int from, to, cap, flow, cost;
 11     Edge(int u, int v, int c, int f, int w): from(u), to(v), cap(c), flow(f), cost(w) {}
 12 };
 13 
 14 struct MCMF
 15 {
 16     int n, m;
 17     vector<Edge> edges;
 18     vector<int> G[maxn];
 19     int inq[maxn]; //是否在队列中
 20     int d[maxn];    //Bellman-Ford
 21     int p[maxn];    //上一条弧
 22     int a[maxn];    //可改进量
 23 
 24     void Init(int n)
 25     {
 26         this->n = n;
 27         for(int i = 0; i < n; ++i) G[i].clear();
 28         edges.clear();
 29     }
 30 
 31     void AddEdge(int from, int to, int cap, int cost)
 32     {
 33         edges.push_back(Edge(from, to, cap, 0, cost));
 34         edges.push_back(Edge(to, from, 0, 0, -cost));
 35         m = edges.size();
 36         G[from].push_back(m-2);
 37         G[to].push_back(m-1);
 38     }
 39 
 40     bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost)
 41     {
 42         for(int i = 0; i < n; ++i) d[i] = INF;
 43         memset(inq, 0, sizeof(inq));
 44         d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
 45 
 46         queue<int> Q;
 47         Q.push(s);
 48         while(!Q.empty())
 49         {
 50             int u = Q.front(); Q.pop();
 51             inq[u] = 0;
 52             for(int i = 0; i < G[u].size(); ++i)
 53             {
 54                 Edge& e = edges[G[u][i]];
 55                 if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
 56                 {
 57                     d[e.to] = d[u] + e.cost;
 58                     p[e.to] = G[u][i];
 59                     a[e.to] = min(a[u], e.cap - e.flow);
 60                     if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
 61                 }
 62             }
 63         }
 64         if(d[t] == INF) return false;
 65         if(flow + a[t] > flow_limit) a[t] = flow_limit - flow;
 66         flow += a[t];
 67         cost += d[t] * a[t];
 68         for(int u = t; u != s; u = edges[p[u]].from)
 69         {
 70             edges[p[u]].flow += a[t];
 71             edges[p[u]^1].flow -= a[t];
 72         }
 73         return true;
 74     }
 75 
 76     int MincostMaxflow(int s, int t, int flow_limit, int& cost)
 77     {
 78         int flow = 0; cost = 0;
 79         while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
 80         return flow;
 81     }
 82 }g;
 83 
 84 int main()
 85 {
 86     //freopen("in.txt", "r", stdin);
 87 
 88     int n, m;
 89     while(scanf("%d%d", &n, &m) == 2 && n)
 90     {
 91         g.Init(n*2-2);
 92         //2~n-1 i和i'的编号分别为1~n-2  n~2n-3
 93         for(int i = 2; i <= n-1; ++i) g.AddEdge(i-1, n-2+i, 1, 0);
 94         for(int i = 0; i < m; ++i)
 95         {   //连接a'->b
 96             int a, b, c;
 97             scanf("%d%d%d", &a, &b, &c);
 98             if(a != 1 && a != n) a += n-2; else a--;
 99             b--;
100             g.AddEdge(a, b, 1, c);
101         }
102         int cost;
103         g.MincostMaxflow(0, n-1, 2, cost);
104         printf("%d
", cost);
105     }
106 
107     return 0;
108 }

代码君

UVa 1658 (拆点法 最小费用流) Admiral

UVa 1658 (拆点法最小费用流) Admiral