Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P2756 飞行员配对方案问题

P2756 飞行员配对方案问题

直接二分图的最大匹配即可。

如何求具体方案?连接两部的边如果有流量,则为匹配。

查看代码
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#include <iostream>
#include <cstdio>
#include <queue>
#include <climits>
#define INF INT_MAX
using namespace std;
const int N = 110, M = 1e4 + 10;
int n, m, st, ed, d[N], cur[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
bool bfs()
{
queue <int> q;
for (int i = 0; i <= n + 1; i++)
d[i] = -1;
q.push(st);
d[st] = 0;
cur[st] = hd[st];
while (!q.empty())
{
int t = q.front();
q.pop();
for (int i = hd[t]; ~i; i = nxt[i])
{
int to = edg[i];
if (d[to] == -1 && wt[i])
{
cur[to] = hd[to];
d[to] = d[t] + 1;
if (to == ed)
return true;
q.push(to);
}
}
}
return false;
}
int dfs(int x, int limit)
{
if (x == ed)
return limit;
int res = 0;
for (int i = cur[x]; ~i && res < limit; i = nxt[i])
{
cur[x] = i;
int to = edg[i];
if (d[to] == d[x] + 1 && wt[i])
{
int t = dfs(to, min(limit - res, wt[i]));
if (!t)
d[to] = -1;
res += t;
wt[i] -= t;
wt[i ^ 1] += t;
}
}
return res;
}
int dinic()
{
int flow, res = 0;
while (bfs())
while (flow = dfs(st, INF))
res += flow;
return res;
}
void add(int x, int y, int z)
{
nxt[++idx] = hd[x];
hd[x] = idx;
edg[idx] = y;
wt[idx] = z;
}
int main()
{
cin >> m >> n;
for (int i = 0; i <= n + 1; i++)
hd[i] = -1;
st = 0;
ed = n + 1;
for (int i = 1; i <= m; i++)
{
add(st, i, 1);
add(i, st, 0);
}
for (int i = m + 1; i <= n; i++)
{
add(i, ed, 1);
add(ed, i, 0);
}
for (int a = 0, b = 0; a != -1 && b != -1; )
{
cin >> a >> b;
add(a, b, 1);
add(b, a, 0);
}
cout << dinic() << endl;
for (int i = 0; i <= idx; i += 2)
if (edg[i] > m && edg[i] <= n && !wt[i])
cout << edg[i ^ 1] << ' ' << edg[i] << endl;
return 0;
}