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

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dinic

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#include <queue>
using namespace std;
const int N = + 10, M = + 10, inf = 1e9;
int n, st, ed, d[N], cur[N];
int idx, hd[N], nxt[M], edg[M], wt[M];
bool bfs ()
{
for (int i = st; i <= ed; ++i) d[i] = -1;
d[st] = 0;
cur[st] = hd[st];
queue <int> q;
q.push(st);
while (!q.empty())
{
int t = q.front(); q.pop();
for (int i = hd[t]; ~i; i = nxt[i])
if (!~d[edg[i]] && wt[i])
{
cur[edg[i]] = hd[edg[i]];
d[edg[i]] = d[t] + 1;
if (edg[i] == ed) return true;
q.push(edg[i]);
}
}
return false;
}
int dfs (int x, int k)
{
if (x == ed) return k;
int res = 0;
for (int i = cur[x]; ~i && res < k; i = nxt[i])
{
cur[x] = i;
if (d[edg[i]] == d[x] + 1 && wt[i])
{
int t = dfs(edg[i], min(wt[i], k - res));
if (!t) d[edg[i]] = -1;
res += t;
wt[i] -= t, wt[i ^ 1] += t;
}
}
return res;
}
int dinic ()
{
int res = 0, flow;
while (bfs()) while (flow = dfs(st, inf)) res += flow;
return res;
}
void add (int a, int b, int c)
{ nxt[idx] = hd[a], hd[a] = idx, edg[idx] = b, wt[idx++] = c; }