1 int cap
[MAXN
+1][MAXN
+1], flow
[MAXN
+1][MAXN
+1], prev
[MAXN
+1];
4 cap[i][j] = Capacidad de la arista (i, j).
5 flow[i][j] = Flujo actual de i a j.
6 prev[i] = Predecesor del nodo i en un camino de aumentaciĆ³n.
9 int fordFulkerson(int n
, int s
, int t
){
11 for (int i
=0; i
<n
; ++i
) fill(flow
[i
], flow
[i
]+n
, 0);
13 fill(prev
, prev
+n
, -1);
16 while (q
.size() && prev
[t
] == -1){
19 for (int v
= 0; v
<n
; ++v
)
20 if ( v
!= s
&& prev
[v
] == -1 && cap
[u
][v
] > flow
[u
][v
] )
21 prev
[v
] = u
, q
.push(v
);
24 if (prev
[t
] == -1) break;
26 int bottleneck
= INT_MAX
;
27 for (int v
= t
, u
= prev
[v
]; u
!= -1; v
= u
, u
= prev
[v
]){
28 bottleneck
= min(bottleneck
, cap
[u
][v
] - flow
[u
][v
]);
30 for (int v
= t
, u
= prev
[v
]; u
!= -1; v
= u
, u
= prev
[v
]){
31 flow
[u
][v
] += bottleneck
;
32 flow
[v
][u
] = -flow
[u
][v
];