最大流模板

最简单的Ford-Fulkerson，复杂度O(FE) （F是最大流量，E是边数

每次从源点到汇点dfs寻找增广路。

const int MAXV = 2005;
const int INF = 1<<30;
struct Edge{ int to, cap, rev; };
std::vector<Edge> G[MAXV];
bool used[MAXV];

void addedge(int from, int to, int cap) {
    G[from].push_back(Edge{to, cap, G[to].size()});
    G[to].push_back(Edge{from, 0, G[from].size()-1});
}

int dfs(int v, int t, int f) {
    if (v == t) return f;
    used[v] = true;
    for (int i = 0; i < G[v].size(); ++i) {
        Edge &e = G[v][i];
        if (!used[e.to] && e.cap > 0) {
            int d = dfs(e.to, t, std::min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

int maxflow(int s, int t) {
    int flow = 0;
    for (; ;) {
        memset(used, 0, sizeof used);
        int f = dfs(s, t, INF);
        if (!f) return flow;
        flow += f;
    }
    return flow;
}

把dfs换成bfs，就成了Edmonds-Karp

代码多了一些，但是并不会快的样子。。

#include <queue>
#include <cstring>
const int N = 2005;
const int M = 2005;
const int INF = 0x7fffffff;

struct Edge {
    int from, to, next, cost;
} edge[M];
int head[N];
int cnt_edge;
void add_edge(int u, int v, int c)
{
    edge[cnt_edge].to = v;
    edge[cnt_edge].from = u;
    edge[cnt_edge].cost = c;
    edge[cnt_edge].next = head[u];
    head[u] = cnt_edge++;
}

int pre[N], flow[N];
std::queue<int> q;
int bfs(int src, int des)
{
    memset(pre, -1, sizeof pre);
    while (!q.empty()) q.pop();
    q.push(src);
    flow[src] = INF;
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        if (u == des) break;
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            int cost = edge[i].cost;
            if (pre[v] == -1 && cost > 0)
            {
                flow[v] = std::min(flow[u], cost);
                pre[v] = i; // 记录的是边
                q.push(v);
            }
        }
    }
    if (pre[des] == -1) return -1;
    return flow[des];
}

int maxFlow(int src, int des)
{
    int ans = 0;
    int in;
    while ((in = bfs(src, des)) != -1)
    {
        int k = des;
        while (k != src)
        {
            int last = pre[k];
            edge[last].cost -= in;
            edge[last ^ 1].cost += in;
            k = edge[last].from;
        }
        ans += in;
    }
    return ans;
}

int main()
{
    int n, m;
    while (~scanf("%d%d", &m, &n))
    {
        int a, b, c;
        memset(head, -1, sizeof head);
        while (m--)
        {
            scanf("%d%d%d", &a, &b, &c);
            add_edge(a, b, c);
            add_edge(b, a, 0);
        }
        printf("%d\n", maxFlow(1, n));
    }
}

优化下，bfs构造分层图，然后每次都走最短的增广路，变成Dinic

这样比上面快很多。。复杂度O(EV²) （E是边数，V是点数

有了大概这个可以放弃上面的两个了。。

据说比较适合有向无环图。。

#include <cstdio>
#include <vector>
#include <cstring>
#include <queue>
const int MAXV = 2005;
const int INF = 1<<30;
struct Edge{ int to, cap, rev; };
std::vector<Edge> G[MAXV];
int level[MAXV];
int iter[MAXV]; //当前弧，之前的已经没有用了

void addedge(int from, int to, int cap) {
    G[from].push_back(Edge{to, cap, G[to].size()});
    G[to].push_back(Edge{from, 0, G[from].size()-1});
}

void bfs(int s) {
    memset(level, -1, sizeof level);
    std::queue<int> que;
    level[s] = 0;
    que.push(s);
    while (!que.empty()) {
        int v = que.front(); que.pop();
        for (int i = 0; i < G[v].size(); ++i) {
            Edge &e = G[v][i];
            if (e.cap > 0 && level[e.to] < 0) {
                level[e.to] = level[v] + 1;
                que.push(e.to);
            }
        }
    }
}

int dfs(int v, int t, int f) {
    if (v == t) return f;
    for (int &i = iter[v]; i < G[v].size(); ++i) { // 注意i是引用 实现当前弧优化
        Edge &e = G[v][i];
        if (e.cap > 0 && level[v] < level[e.to]) {
            int d = dfs(e.to, t, std::min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

int maxflow(int s, int t) {
    int flow = 0;
    for (; ;) {
        bfs(s);
        if (level[t] < 0) return flow;
        memset(iter, 0, sizeof iter);
        int f;
        while ((f = dfs(s, t, INF)) > 0) {
            flow += f;
        }
    }
    return flow;
}

SAP。。。并不是很懂。。。

貌似没有比dinic快很多。。。

const int N = 1000;
const int M = 1000000;
const int INF = 1 << 30;
struct Edge {
    int from, to, next, w;//from一般用不到
} edge[M];
int head[N], cntE;
int src, sink;
int pre[N], cur[N], dis[N], gap[N];
int que[N], open, tail;

void addedge(int u, int v, int w) {
    edge[cntE].from = u;
    edge[cntE].to = v;
    edge[cntE].w = w;
    edge[cntE].next = head[u];
    head[u] = cntE++;
    edge[cntE].from = v;
    edge[cntE].to = u;
    edge[cntE].w = 0;
    edge[cntE].next = head[v];
    head[v] = cntE++;
}
void BFS() {
    int i, u, v;
    memset(gap, 0, sizeof(gap));
    memset(dis, -1, sizeof(dis));
    open = tail = 0;
    que[open] = sink;
    dis[sink] = 0;
    while (open <= tail) {
        u = que[open++];
        for (i = head[u]; ~i; i = edge[i].next) {
            v = edge[i].to;
            if (edge[i].w != 0 || dis[v] != -1) continue;
            que[++tail] = v;
            ++gap[dis[v] = dis[u] + 1];
        }
    }
}
int sap(int n) { //编号从1开始 1~n
    int i, v, u, flow = 0, aug = INF;
    int flag;
    BFS();
    gap[0] = 1;
    for (i = 1; i <= n; i++) cur[i] = head[i];
    u = pre[src] = src;
    while (dis[src] < n) {
        flag = 0;
        for (int j = cur[u]; ~j; j = edge[j].next) {
            v = edge[j].to;
            if (edge[j].w > 0 && dis[u] == dis[v] + 1) {
                flag = 1;
                if (edge[j].w < aug) aug = edge[j].w;
                pre[v] = u; u = v;
                if (u == sink) {
                    flow += aug;
                    while (u != src) {
                        u = pre[u];
                        edge[cur[u]].w -= aug;
                        edge[cur[u] ^ 1].w += aug;
                    }
                    aug = INF;
                }
                break;
            }
            cur[u] = edge[j].next;
        }
        if (flag) continue;
        int mindis = n;
        for (int j = head[u]; ~j; j = edge[j].next) {
            v = edge[j].to;
            if (edge[j].w > 0 && mindis > dis[v]) {
                mindis = dis[v];
                cur[u] = j;
            }
        }
        if (--gap[dis[u]] == 0) break;
        ++gap[dis[u] = mindis + 1];
        u = pre[u];
    }
    return flow;
}

int main() {
    memset(head, -1, sizeof head);
    cntE = 0;
}

我决定选择dinic吧。。。好写。。。Orz。。