[P2756] 飞行员配对方案问题 - 二分图匹配
Description
一共有 n 个飞行员，其中有 m 个外籍飞行员和 (n−m) 个英国飞行员。对于给定的外籍飞行员与英国飞行员的配合情况，试设计一个算法找出最佳飞行员配对方案，使皇家空军一次能派出最多的飞机。输出方案。
Solution
题目是简单的二分图匹配，今天的主题是怎么魔改封装好的板子来输出方案。
我们给 make 函数增加一个返回值，表示这条边的正向边的编号，新增一个 get_value 函数，用于获取某条边上的流量
于是在建图的过程中，在连接跨部边时，我们将边的编号和对应的飞行员的映射关系记下来（用边的编号索引），最后查表即可
#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define int long long
#ifndef __FLOW_HPP__
#define __FLOW_HPP__
// v1.1 feat. edge query for Maxf
#include <bits/stdc++.h>
using namespace std;
#define int long long
namespace flowsolution
{
const int N = 100005;
const int M = 1000005;
const int inf = 1e+12;
struct MaxflowSolution
{
int *dis, ans, cnt = 1, s, t, *pre, *next, *head, *val;
MaxflowSolution()
{
cnt = 1;
dis = new int[N];
pre = new int[M];
next = new int[M];
head = new int[N];
val = new int[M];
fill(dis, dis + N, 0);
fill(pre, pre + M, 0);
fill(next, next + M, 0);
fill(head, head + N, 0);
fill(val, val + M, 0);
}
~MaxflowSolution()
{
delete[] dis;
delete[] pre;
delete[] next;
delete[] head;
delete[] val;
}
std::queue<int> q;
int make(int x, int y, int z)
{
pre[++cnt] = y, next[cnt] = head[x], head[x] = cnt, val[cnt] = z;
int ret = cnt;
pre[++cnt] = x, next[cnt] = head[y], head[y] = cnt;
return ret;
}
int get_value(int x)
{
return val[x];
}
bool bfs()
{
fill(dis, dis + N, 0);
q.push(s), dis[s] = 1;
while (!q.empty())
{
int x = q.front();
q.pop();
for (int i = head[x]; i; i = next[i])
if (!dis[pre[i]] && val[i])
dis[pre[i]] = dis[x] + 1, q.push(pre[i]);
}
return dis[t];
}
int dfs(int x, int flow)
{
if (x == t || !flow)
return flow;
int f = flow;
for (int i = head[x]; i; i = next[i])
if (val[i] && dis[pre[i]] > dis[x])
{
int y = dfs(pre[i], min(val[i], f));
f -= y, val[i] -= y, val[i ^ 1] += y;
if (!f)
return flow;
}
if (f == flow)
dis[x] = -1;
return flow - f;
}
int solve(int _s, int _t)
{
s = _s;
t = _t;
ans = 0;
for (; bfs(); ans += dfs(s, inf))
;
return ans;
}
};
struct CostflowSolution
{
struct Edge
{
int p = 0, c = 0, w = 0, next = -1;
} * e;
int s, t, tans, ans, cost, ind, *bus, qhead = 0, qtail = -1, *qu, *vis, *dist;
CostflowSolution()
{
e = new Edge[M];
qu = new int[M];
bus = new int[N];
vis = new int[N];
dist = new int[N];
fill(qu, qu + M, 0);
fill(bus, bus + N, -1);
fill(vis, vis + N, 0);
fill(dist, dist + N, 0);
ind = 0;
}
~CostflowSolution()
{
delete[] e;
delete[] qu;
delete[] vis;
delete[] dist;
}
void graph_link(int p, int q, int c, int w)
{
e[ind].p = q;
e[ind].c = c;
e[ind].w = w;
e[ind].next = bus[p];
bus[p] = ind;
++ind;
}
void make(int p, int q, int c, int w)
{
graph_link(p, q, c, w);
graph_link(q, p, 0, -w);
}
int dinic_spfa()
{
qhead = 0;
qtail = -1;
fill(vis, vis + N, 0);
fill(dist, dist + N, inf);
vis[s] = 1;
dist[s] = 0;
qu[++qtail] = s;
while (qtail >= qhead)
{
int p = qu[qhead++];
vis[p] = 0;
for (int i = bus[p]; i != -1; i = e[i].next)
if (dist[e[i].p] > dist[p] + e[i].w && e[i].c > 0)
{
dist[e[i].p] = dist[p] + e[i].w;
if (vis[e[i].p] == 0)
vis[e[i].p] = 1, qu[++qtail] = e[i].p;
}
}
return dist[t] < inf;
}
int dinic_dfs(int p, int lim)
{
if (p == t)
return lim;
vis[p] = 1;
int ret = 0;
for (int i = bus[p]; i != -1; i = e[i].next)
{
int q = e[i].p;
if (e[i].c > 0 && dist[q] == dist[p] + e[i].w && vis[q] == 0)
{
int res = dinic_dfs(q, min(lim, e[i].c));
cost += res * e[i].w;
e[i].c -= res;
e[i ^ 1].c += res;
ret += res;
lim -= res;
if (lim == 0)
break;
}
}
return ret;
}
pair<int, int> solve(int _s, int _t)
{
s = _s;
t = _t;
ans = 0;
cost = 0;
while (dinic_spfa())
{
fill(vis, vis + N, 0);
ans += dinic_dfs(s, inf);
}
return make_pair(ans, cost);
}
};
} // namespace flowsolution
#endif
signed main()
{
ios::sync_with_stdio(false);
int n, m;
cin >> m >> n;
int n1 = m, n2 = n - m;
flowsolution::MaxflowSolution flow;
unordered_map<int, pair<int, int>> mp;
int s = n + 1, t = n + 2;
for (int i = 1; i <= n1; i++)
flow.make(s, i, 1);
for (int i = n1 + 1; i <= n; i++)
flow.make(i, t, 1);
int t1, t2;
while (cin >> t1 >> t2)
{
if (t1 == -1)
break;
int eid = flow.make(t1, t2, 1);
mp[eid] = {t1, t2};
}
cout << flow.solve(s, t) << endl;
for (auto [eid, pr] : mp)
{
if (flow.get_value(eid ^ 1))
{
cout << pr.first << " " << pr.second << endl;
}
}
}