UVa1658 Admiral (拆点法,最小费用流)

Posted 2020-08-10

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 }