代碼中求的是最小費用最大流，求最大費用最大流只需要在設置邊權時設置為原權值的相反數，執行一次最小費用最大流，計算得出最小費用的相反數就是要求的最大費用。

u[x], v[x], w[x], c[x] 分別表示 x 號邊的出發點，到達點，權值和殘量。

 1 int SPFA()
 2 {
 3     int head, tail, i, node, p;
 4     for (i = 1; i <= ver; ++i) {
 5         dis[i] = INF;
 6         path[i] = 0;
 7         bok[i] = false;
 8     }
 
 9     dis[s] = 0;
10     head = tail = 1;
11     que[tail++] = s;
12     bok[s] = true;
13     while (head < tail) {
14         node = que[head++];
15         bok[node] = false;
16         for (p = fst[node]; p; p = nxt[p]) {
17             if (c[p] <= 0 || dis[node] + w[p] >= dis[v[p]])
18                 continue
 
;
19             dis[v[p]] = dis[node] + w[p];
20             path[v[p]] = p
21             if (bok[v[p]])
22                 continue;
23             que[tail++] = v[p];
24             bok[v[p]] = true;
25         }
26     }
27     return dis[t];
28 }
29 void modify()
30 {
31     int tmp = INF;
32     for
 
 (p = path[t]; p; p = path[u[p]])
33         tmp = min(c[p], tmp);
34     cost += tmp * dis[t];
35     maxflow += tmp;
36     for (p = path[t]; p; p = path[u[p]]) {
37         c[p] -= tmp;
38         c[p + 1] += tmp;
39     }
40     return ;
41 }
42 
43 int main()
44 {
45     /* input */
46     while (SPFA() < INF)
47         modify();
48     /* now we get ‘cost‘ and ‘maxflow‘ */
49     return 0;
50 }

費用流偽代碼