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UVa1658 Admiral (拆点法,最小费用流)

链接:http://vjudge.net/problem/UVA-1658

 

分析:把2到v-1的每个节点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。

 1 #include <cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<vector> 5 #include<algorithm> 6 using namespace std; 7  8 const int maxn = 2000 + 5; 9 const int INF = 1000000000;10 11 struct Edge {12     int from, to, cap, flow, cost;13     Edge(int u, int v, int c, int f, int w):from(u), to(v), cap(c), flow(f), cost(w) {}14 };15 16 struct MCMF {17     int n, m;18     vector<Edge> edges;19     vector<int> G[maxn];20     int inq[maxn];21     int d[maxn];22     int p[maxn];23     int a[maxn];24 25     void init(int n) {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         edges.push_back(Edge(from, to, cap, 0, cost));33         edges.push_back(Edge(to, from, 0, 0, -cost));34         m = edges.size();35         G[from].push_back(m - 2);36         G[to].push_back(m - 1);37     }38 39     bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost) {40         for (int i = 0; i < n; i++) d[i] = INF;41         memset(inq, 0, sizeof(inq));42         d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;43 44         queue<int> Q;45         Q.push(s);46         while (!Q.empty()) {47             int u = Q.front(); Q.pop();48             inq[u] = 0;49             for (int i = 0; i < G[u].size(); i++) {50                 Edge& e = edges[G[u][i]];51                 if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {52                     d[e.to] = d[u] + e.cost;53                     p[e.to] = G[u][i];54                     a[e.to] = min(a[u], e.cap - e.flow);55                     if (!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }56                 }57             }58         }59         if (d[t] == INF) return false;60         if (flow + a[t] > flow_limit) a[t] = flow_limit - flow;61         flow += a[t];62         cost += d[t] * a[t];63         for (int u = t; u != s; u = edges[p[u]].from) {64             edges[p[u]].flow += a[t];65             edges[p[u] ^ 1].flow -= a[t];66         }67         return true;68     }69 70     int MincostFlow(int s, int t, int flow_limit, int& cost) {71         int flow = 0; cost = 0;72         while (flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));73         return flow;74     }75 };76 77 MCMF g;78 79 int main() {80     int v, e;81     while (scanf("%d%d", &v, &e) == 2 && v) {82         g.init(v * 2 - 2);83         for (int i = 2; i <= v - 1; i++)84             g.AddEdge(i - 1, v + i - 2, 1, 0);85         int a, b, c;86         while (e--) {87             scanf("%d%d%d", &a, &b, &c);88             if (a != 1 && a != v) a += v - 2; else a--;89             b--;90             g.AddEdge(a, b, 1, c);91         }92         int cost;93         g.MincostFlow(0, v - 1, 2, cost);94         printf("%d\n", cost);95     }96     return 0;97 }

 

UVa1658 Admiral (拆点法,最小费用流)