P2756 飞行员配对方案问题 网络流

P2756 飞行员配对方案问题

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 10005, inf = 0x3f3f3f;
 4 struct Edge {
 5     int from, to, cap, flow;
 6 };
 7 
 8 struct Dinic {
 9     int n, m, s, t;
10     vector<Edge> edges;
11     vector<int> G[maxn];
12     bool vis[maxn];
13     int d[maxn];
14     int cur[maxn];
15 
16     void AddEdge(int from, int to, int cap) {
17         edges.push_back((Edge){from, to, cap, 0});
18         edges.push_back((Edge){to, from, 0, 0});
19         m = edges.size();
20         G[from].push_back(m-2);
21         G[to].push_back(m-1);
22     }
23     bool bfs() {
24         memset(vis, 0, sizeof(vis));
25         queue<int> que;
26         que.push(s);
27         d[s] = 0;
28         vis[s] = true;
29         while (!que.empty()) {
30             int x = que.front(); que.pop();
31             for (int i = 0; i < G[x].size(); ++i) {
32                 Edge& e = edges[G[x][i]];
33                 if (!vis[e.to] && e.cap > e.flow) {
34                     vis[e.to] = true;
35                     d[e.to] = d[x] + 1;
36                     que.push(e.to);
37                 }
38             }
39         }
40         return vis[t];
41     }
42     int dfs(int x, int a) {
43         if (x == t || a == 0) return a;
44         int flow = 0, f;
45         for (int& i = cur[x]; i < G[x].size(); ++i) {
46             Edge& e = edges[G[x][i]];
47             if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0) {
48                 e.flow += f;
49                 edges[G[x][i]^1].flow -= f;
50                 flow += f;
51                 a -= f;
52                 if (a == 0) break;
53             }
54         }
55         return flow;
56     }
57     int maxflow(int s, int t) {
58         this->s = s; this->t = t;
59         int flow = 0;
60         while (bfs()) {
61             memset(cur,0,sizeof(cur));
62             flow += dfs(s,inf);
63         }
64         return flow;
65     }
66 }dinic;
67 int main() {
68     int m, n; scanf("%d%d",&m,&n);
69     while (true) {
70         int i, j; scanf("%d%d",&i,&j);
71         if (i == -1 && j == -1) break;
72         dinic.AddEdge(i,j,1);
73     }
74     int s = n+1, t = n+2;
75     for (int i = 1; i <= m; ++i) {
76         dinic.AddEdge(s,i,1);
77     }
78     for (int i = m+1; i <= n; ++i) {
79         dinic.AddEdge(i,t,1);
80     }
81     int ans = dinic.maxflow(s,t);
82     if (ans == 0) puts("No Solution!");
83     else {
84         printf("%d\n",ans);
85         for (int i = 0; i < dinic.edges.size(); ++i) {
86             if (dinic.edges[i].from == s) continue;
87             if (dinic.edges[i].to == t) continue;
88             if (dinic.edges[i].flow == 1) {
89                 printf("%d %d\n",dinic.edges[i].from,dinic.edges[i].to);
90             }
91         }
92     }
93     return 0;
94 }

 

posted @ 2019-11-05 13:43  麻辣猪仔  阅读(140)  评论(0)    收藏  举报