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 }
