最大流模板

 

这个ISAP模板只能增广一次，在残量网络上跑就会有问题，但是效率不错。

 1 namespace ISAP {
 2     int tot, n, m, src, tar, qh, qt, cnt;
 3     ll ans;
 4     struct edge {
 5         int vet, next, len;
 6     } E[LEN * 2];
 7     int dis[LEN], gap[LEN], head[LEN], cur[LEN], q[LEN], vis[LEN];
 8     void add(int u, int v, int c) {
 9         E[++tot] = (edge){v, head[u], c};
10         head[u] = tot;
11     }
12     void join(int u, int v, int c) {
13         add(u, v, c);
14         add(v, u, 0);
15     }
16     void bfs(int s) {
17         qh = qt = 0; 
18         q[++qt] = s;
19         dis[s] = 0;
20         vis[s] = 1;
21         while (qh < qt) {
22             int u = q[++qh];
23             gap[dis[u]]++;
24             for (int e = head[u]; e != -1; e = E[e].next) {
25                 int v = E[e].vet;
26                 if (E[e ^ 1].len && !vis[v]) {
27                     dis[v] = dis[u] + 1;
28                     vis[v] = 1;
29                     q[++qt] = v;
30                 }
31             }
32         }
33     }
34     int isap(int u, int aug) {
35         if (u == tar) return aug;
36         int flow = 0;
37         for (int e = head[u]; e != -1; e = E[e].next) {
38             int v = E[e].vet;
39             if (E[e].len && dis[v] == dis[u] - 1) {
40                 int tmp = isap(v, min(aug - flow, E[e].len));
41                 E[e].len -= tmp;
42                 E[e ^ 1].len += tmp;
43                 flow += tmp;
44                 head[u] = e;
45                 if (flow == aug || dis[src] == cnt) return flow;
46             }
47         }
48         if (!--gap[dis[u]++]) dis[src] = cnt;
49         ++gap[dis[u]];
50         head[u] = cur[u];
51         return flow;
52     }
53     ll maxflow(int s, int t) {
54         ll res = 0;
55         src = s, tar = t;
56         for (int i = 1; i <= cnt; i++) cur[i] = head[i];
57         bfs(tar);
58         while (dis[src] < cnt) res += isap(src, oo);
59         return res;
60     }
61     void init() {
62         tot = -1, gap[0] = 0;
63         for (int i = 1; i <= cnt; i++) {
64             vis[i] = gap[i] = dis[i] = 0;
65             head[i] = -1;
66         }
67     }
68 }
ISAP

这个DINIC模板可以在残量网络上增广多次。

 1 namespace DINIC {
 2     int tot, src, tar, n, m, qh, qt, cnt;
 3     int head[LEN], cur[LEN], dis[LEN], q[LEN];
 4     struct edge {
 5         int vet, next, len;
 6     } E[LEN * 2];
 7     void add(int u, int v, int c) {
 8         E[++tot] = (edge){v, head[u], c};
 9         head[u] = tot;
10     }
11     void join(int u, int v, int c) {
12         add(u, v, c);
13         add(v, u, 0);
14     }
15     void init() {
16         tot = -1;
17         for (int i = 1; i <= cnt; i++) head[i] = -1;
18     }
19     bool bfs() {
20         for (int i = 1; i <= cnt; i++) dis[i] = 0;
21         qh = qt = 0;
22         q[++qt] = src;
23         dis[src] = 1;
24         while (qh < qt) {
25             int u = q[++qh];
26             for (int e = head[u]; e != -1; e = E[e].next) {
27                 int v = E[e].vet;
28                 if (E[e].len && !dis[v]) {
29                     dis[v] = dis[u] + 1;
30                     if (v == tar) return 1;
31                     q[++qt] = v;
32                 }
33             }
34         }
35         return dis[tar];
36     }
37     int dfs(int u, int aug) {
38         if (u == tar || !aug) return aug;
39         int tmp = 0;
40         for (int &e = cur[u]; e != -1; e = E[e].next) {
41             int v = E[e].vet;
42             if (dis[v] == dis[u] + 1) {
43                 if (tmp = dfs(v, min(aug, E[e].len))) {
44                     E[e].len -= tmp;
45                     E[e ^ 1].len += tmp;
46                     return tmp;
47                 }
48             }
49         }
50         return 0;
51     }
52     ll maxflow(int s, int t) {
53         src = s, tar = t;
54         ll res = 0;
55         int flow = 0;
56         while (bfs()) {
57             for (int i = 1; i <= cnt; i++) cur[i] = head[i];
58             while (flow = dfs(src, oo)) res += flow;
59         }
60         return res;
61     }
62 }
DINIC

 

posted on 2018-08-03 18:52  NineSwords  阅读(223)  评论(0)  编辑  收藏  举报

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