「网络流浅谈」最大流的应用

二分图匹配

考虑如何将二分图匹配问题，转化为流网络。设置 $1$ 个汇点和源点，从源点向二分图一侧的每一个点连边，从另一侧向汇点连边，边权均为 $1$，二分图中的边也全部加入，权值设为 $1$。这样，二分图的最大匹配等于流网络的最大流。

P2756 飞行员配对方案问题

题意：给定 $1$ 个二分图，求最大匹配。

匈牙利算法是可以求二分图最大匹配的，不过太慢了。不妨，使用上述的方式建立出流网络，并使用 Dinic 求解出该网络的最大流即可。

举个例子：左图为样例的二分图，而右图为建立的流网络。

浅析流网络最大流与二分图最大匹配的相等性：

对于网络流的题目，只需要考虑对于任意的一个最大匹配，都能对应到一个可行流；而对于任意一个可行流都能对应到一个最大匹配。

对于任意的一个最大匹配，都能对应到一个可行流：若选择边 $E_1,E_2,\dots ,E_k$，则可行流中的这些边均为 $1$，且令这些边左端的顶点分别为 $V_1,V_2,\dots ,V_t$，右端的为 $V_1^{\prime },V_2^{\prime },\dots ,V_t^{\prime }$，则可行流的 $s\to V_i$ 这些边均为 $1$；$V_i^{\prime }\to t$ 这些边也均为 $1$。由于匹配不存在 $2$ 条边有公共顶点，所以一定满足容量限制与流量守恒。

对于任意的一个可行流，都能对应到一个最大匹配：可行流中流量为 $1$ 的没有 $s$ 和 $t$ 的边即为最大匹配，由于流量守恒，最多有 $1$ 条边流向一个点，所以满足对于任意 $2$ 条边，都不存在公共点。

故，只需要用 Dinic 跑一遍最大流即可，输出方案就是找出所有反向边流量为 $1$（或正向边流量为 $0$）的边即可。

注意：二分图下的 Dinic 算法极为特殊，时间复杂度为 $O(n^2\sqrt{n})$

#include 
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 1e2 + 10, M = 1e5 + 10;

int n, m, s, t;
int h[N], e[M], ne[M], f[M], idx;
int d[N], cur[N];

void add(int a, int b, int c) {
	e[idx] = b, ne[idx] = h[a], f[idx] = c, h[a] = idx ++;
	e[idx] = a, ne[idx] = h[b], f[idx] = 0, h[b] = idx ++;
}
bool bfs() {
	memset(d, -1, sizeof d);
	queue<int> q;
	q.emplace(s), cur[s] = h[s], d[s] = 0;
	while (q.size()) {
		auto u = q.front();
		q.pop();

		for (int i = h[u]; ~i; i = ne[i]) {
			int j = e[i];
			if (d[j] == -1 && f[i]) {
				d[j] = d[u] + 1, cur[j] = h[j];
				if (j == t) return 1;
				q.emplace(j);
			}
		}
	}
	return 0;
}
int find(int u, int lim) {
	if (u == t) return lim;

	int flow = 0;
	for (int i = cur[u]; ~i && flow < lim; i = ne[i]) {
		int j = e[i];
		if (d[j] == d[u] + 1 && f[i]) {
			int tmp = find(j, min(lim - flow, f[i]));
			if (!tmp) d[j] = -1;
			f[i] -= tmp, f[i ^ 1] += tmp, flow += tmp;
		}
	}
	return flow;
}
int dinic() {
	int res = 0, flow;
	while (bfs()) while (flow = find(s, 1e18)) res += flow;
	return res;
}

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	memset(h, -1, sizeof h);

	cin >> m >> n;

	s = 0, t = n + 1;
	int u, v;
	while (cin >> u >> v && u != -1) {
		add(u, v, 1);
	}
	for (int i = 1; i <= m; i ++)
		add(s, i, 1);
	for (int i = m + 1; i <= n; i ++)
		add(i, t, 1);

	cout << dinic() << endl;
	for (int i = 0; i < idx; i += 2)
		if (e[i] != t && e[i ^ 1] != s && !f[i])
			cout << e[i ^ 1] << " " << e[i] << endl;

	return 0;
}

习题

P3254 圆桌问题，与原建图方式有略微差异。

参考代码

#include 
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 5e2 + 10, M = 1e5 + 10;

int m, n, s, t;
int a[N], b[N];
int h[N], e[M], f[M], ne[M], idx;
int d[N], cur[N];

void add(int a, int b, int c) {
	e[idx] = b, ne[idx] = h[a], f[idx] = c, h[a] = idx ++;
	e[idx] = a, ne[idx] = h[b], f[idx] = 0, h[b] = idx ++;
}
bool bfs() {
	memset(d, -1, sizeof d);
	queue<int> q;
	q.emplace(s), d[s] = 0, cur[s] = h[s];
	while (q.size()) {
		int u = q.front();
		q.pop();

		for (int i = h[u]; ~i; i = ne[i]) {
			int j = e[i];
			if (d[j] == -1 && f[i]) {
				d[j] = d[u] + 1;
				cur[j] = h[j];
				if (j == t) return 1;
				q.emplace(j);
			}
		}
	}
	return 0;
}
int find(int u, int lim) {
	if (u == t) return lim;

	int flow = 0;
	for (int i = cur[u]; ~i && flow < lim; i = ne[i]) {
		cur[u] = i;
		int j = e[i];
		if (d[j] == d[u] + 1 && f[i]) {
			int tmp = find(j, min(lim - flow, f[i]));
			if (!tmp) d[j] = -1;
			f[i] -= tmp, f[i ^ 1] += tmp, flow += tmp;
		}
	}

	return flow;
}
int dinic() {
	int res = 0, flow;
	while (bfs()) while (flow = find(s, 1e18)) res += flow;
	return res;
}

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	memset(h, -1, sizeof h);
	cin >> m >> n;

	s = 0, t = n + m + 1;
	int tot = 0;
	for (int i = 1; i <= m; i ++)
		cin >> a[i], add(s, i, a[i]), tot += a[i];
	for (int i = 1; i <= n; i ++)
		cin >> b[i], add(i + m, t, b[i]);
	for (int i = 1; i <= m; i ++)
		for (int j = m + 1; j <= n + m; j ++)
			add(i, j, 1);

	if (dinic() == tot) {
		cout << 1 << endl;
		std::vector<vector<int>> way(m + 1);
		for (int i = 0; i < idx; i += 2)
			if (e[i] != t && e[i ^ 1] != s && !f[i])
				way[e[i ^ 1]].emplace_back(e[i] - m);
		for (int i = 1; i <= m; i ++) {
			for (auto v : way[i])
				cout << v << " ";
			cout << endl;
		}
	} else {
		cout << 0 << endl;
	}

	return 0;
}

---

多源汇最大流

本质上只不过是源点不是 $1$ 个，汇点也不是 $1$ 个了，那么其实只需要再设一个超级源点连向所有源点，边权为 $+\mathrm{\infty }$，表示向这些源点可以流任意多流量，也就是说从这些源点可以流出任意多流量；同样的，从每一个汇点向超级汇点连一条 $+\mathrm{\infty }$ 的边，表示这些汇点可以流向超级汇点任意多流量，也就是说这些汇点都可以接纳任意多的流量。

这样的新流网络的最大流就是源网络的最大流，所以只需要对于新网络跑一遍 Dinic 即可。

习题

AcWing 2234. 多源汇最大流，模版题

参考代码

#include 
#include 
#include 
#define int long long

using namespace std;

typedef pair<int, int> PII;

const int SIZE = 5e5 + 10;

int N, M, Sc, Tc, S, T;
int h[SIZE], e[SIZE], ne[SIZE], f[SIZE], idx;
int D[SIZE], Current[SIZE];

void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], f[idx] = c, h[a] = idx ++;
    e[idx] = a, ne[idx] = h[b], f[idx] = 0, h[b] = idx ++;
}

bool BFS() {
    memset(D, -1, sizeof D);
    queue<int> Q;

    Q.push(S), D[S] = 0, Current[S] = h[S];
    while (Q.size()) {
        int u = Q.front();
        Q.pop();

        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (D[j] == -1 && f[i]) {
                D[j] = D[u] + 1;
                Current[j] = h[j];
                if (j == T) return true;
                Q.push(j);
            }
        }
    }

    return false;
}

int Find(int u, int limit) {
    if (u == T) return limit;

    int flow = 0;
    for (int i = Current[u]; ~i && flow < limit; i = ne[i]) {
        Current[u] = i;
        int j = e[i];
        if (D[j] == D[u] + 1 && f[i]) {
            int T = Find(j, min(f[i], limit - flow));
            if (!T) D[j] = -1;
            f[i] -= T, f[i ^ 1] += T, flow += T;
        }
    }

    return flow;
}

int Dinic() {
    int Result = 0, flow;
    while (BFS()) while (flow = Find(S, 1e18)) Result += flow;
    return Result;
}

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    memset(h, -1, sizeof h);

    cin >> N >> M >> Sc >> Tc;

    S = 0, T = N + 1;
    while (Sc --) {
        int u;
        cin >> u;
        add(S, u, 1e18);
    }

    while (Tc --) {
        int u;
        cin >> u;
        add(u, T, 1e18);
    }

    while (M --) {
        int a, b, c;

        cin >> a >> b >> c;

        add(a, b, c);
    }

    cout << Dinic() << endl;

    return 0;
}

---

关键边

POJ3204 Ikki's Story I - Road Reconstruction

题意：给定 $1$ 个流网络，求有多少条边，满足增加该边边权后能使最大流增加。

考虑一条边满足什么条件使得增加容量后会使得最大流增加，回顾求最大流的过程：每一次在残留网络中找增广路径，并加到最大流中。

那么，如果容量增加后，最大流增加，那么必然是增加流量后产生 $1$ 条增广路径。所以，对于每一条边 $(u,v)$，只需要判断是否存在 $1$ 条增广路径 $s\to u$ 以及 $1$ 增广路径 $v\to t$。判断的方法就是在最大流的残留网络中 DFS，记录每次走 $>0$ 的边能到达那些点即可。

#include 
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 5e2 + 10, M = 2e4 + 10;

int n, m, s, t;
int h[N], e[M], ne[M], f[M], idx;
int d[N], cur[N], vis[2][N];

void add(int a, int b, int c) {
	e[idx] = b, ne[idx] = h[a], f[idx] = c, h[a] = idx ++;
	e[idx] = a, ne[idx] = h[b], f[idx] = 0, h[b] = idx ++;
}
bool bfs() {
	memset(d, -1, sizeof d);
	queue<int> q;
	q.emplace(s), d[s] = 0, cur[s] = h[s];

	while (q.size()) {
		int u = q.front();
		q.pop();

		for (int i = h[u]; ~i; i = ne[i]) {
			int j = e[i];
			if (d[j] == -1 && f[i]) {
				d[j] = d[u] + 1, cur[j] = h[j];
				if (j == t) return 1;
				q.emplace(j);
			}
		}
	}
	return 0;
}
int find(int u, int lim) {
	if (u == t) return lim;

	int flow = 0;
	for (int i = cur[u]; ~i && flow < lim; i = ne[i]) {
		cur[u] = i;
		int j = e[i];
		if (d[j] == d[u] + 1 && f[i]) {
			int tmp = find(j, min(lim - flow, f[i]));
			if (!tmp) d[j] = -1;
			f[i] -= tmp, f[i ^ 1] += tmp, flow += tmp;
		}
	}

	return flow;
}
int dinic() {
	int res = 0, flow;
	while (bfs()) while (flow = find(s, 1e18)) res += flow;
	return res;
}
void dfs(int u, int k) {
	vis[k][u] = 1;
	for (int i = h[u]; ~i; i = ne[i]) {
		int j = e[i];
		if (!vis[k][j] && f[i ^ k])
			dfs(j, k);
	}
}

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	memset(h, -1, sizeof h);
	cin >> n >> m;

	s = 0, t = n - 1;
	while (m --) {
		int u, v, w;
		cin >> u >> v >> w;
		add(u, v, w);
	}
	dinic();
	dfs(s, 0), dfs(t, 1);

	int res = 0;
	for (int i = 0; i < idx; i += 2)
		if (vis[0][e[i ^ 1]] && vis[1][e[i]])
			res ++;

	cout << res << endl;

	return 0;
}

拆点

POJ3281 Dining

题意：有 $n$ 头奶牛，$F$ 个食物和 $D$ 个饮料，每头奶牛可以吃某些食物和饮料，但都只能吃食物和饮料各一个。求最多能满足多少头奶牛。（三分图匹配）

考虑继续使用类似二分图的建网络流的方式，举个例子：

不过，这样真的能够求出最终的答案吗？答案是否定的。

考虑局部的这样一个位置，最大流得到话会流出 $2$ 的，也就是这个奶牛会贡献 $2$，而应该是 $1$。

所以，就要拆点了！

通过，流量守恒，就可以使得通过每一个点的流量最多为 $1$，也就满足了题意。

#include 
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 4e2 + 10, M = 1e5 + 10;

int n, m, k, s, t;
int h[N], e[M], ne[M], f[M], idx;
int d[N], cur[N];

void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], f[idx] = c, h[a] = idx ++;
    e[idx] = a, ne[idx] = h[b], f[idx] = 0, h[b] = idx ++;
}
bool bfs() {
    memset(d, -1, sizeof d);
    queue<int> q;
    q.emplace(s), d[s] = 0, cur[s] = h[s];
    while (q.size()) {
        int u = q.front();
        q.pop();

        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (d[j] == -1 && f[i]) {
                d[j] = d[u] + 1, cur[j] = h[j];
                if (j == t) return 1;
                q.emplace(j);
            }
        }
    }
    return 0;
}
int find(int u, int lim) {
    if (u == t) return lim;

    int flow = 0;
    for (int i = cur[u]; ~i && flow < lim; i = ne[i]) {
        cur[u] = i;
        int j = e[i];
        if (d[j] == d[u] + 1 && f[i]) {
            int tmp = find(j, min(lim - flow, f[i]));
            if (!tmp) d[j] = -1;
            f[i] -= tmp, f[i ^ 1] += tmp, flow += tmp;
        }
    }
    return flow;
}
int dinic() {
    int res = 0, flow;
    while (bfs()) while (flow = find(s, 1e18)) res += flow;
    return res;
}

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    memset(h, -1, sizeof h);
    cin >> n >> m >> k;

    s = 0, t = n * 2 + m + k + 1;
    for (int i = 1; i <= n; i ++) {
        int cf, cd, x;
        cin >> cf >> cd;
        for (int j = 1; j <= cf; j ++)
            cin >> x, add(x, i + m, 1);
        for (int j = 1; j <= cd; j ++)
            cin >> x, add(i + m + n, x + m + n + n, 1);
    }
    for (int i = 1; i <= m; i ++)
        add(s, i, 1);
    for (int i = m + n * 2 + 1; i < t; i ++)
        add(i, t, 1);
    for (int i = m + 1; i <= m + n; i ++)
        add(i, i + n, 1);

    cout << dinic() << endl;

    return 0;
}

习题

P2766 最长不下降子序列问题

参考代码

#include 
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 1e3 + 10, M = 4e5 + 10;

int n, s, t;
int a[N], dp[N];
int h[N], e[M], ne[M], f[M], idx;
int d[N], cur[N];
void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], f[idx] = c, h[a] = idx ++;
    e[idx] = a, ne[idx] = h[b], f[idx] = 0, h[b] = idx ++;
}
bool bfs() {
    memset(d, -1, sizeof d);
    queue<int> q;
    q.emplace(s), d[s] = 0, cur[s] = h[s];
    while (q.size()) {
        int u = q.front();
        q.pop();

        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (d[j] == -1 && f[i]) {
                d[j] = d[u] + 1, cur[j] = h[j];
                if (j == t) return 1;
                q.emplace(j);
            }
        }
    }
    return 0;
}
int find(int u, int lim) {
    if (u == t) return lim;

    int flow = 0;
    for (int i = cur[u]; ~i && flow < lim; i = ne[i]) {
        cur[u] = i;
        int j = e[i];
        if (d[j] == d[u] + 1 && f[i]) {
            int tmp = find(j, min(lim - flow, f[i]));
            if (!tmp) d[j] = -1;
            f[i] -= tmp, f[i ^ 1] += tmp, flow += tmp;
        }
    }
    return flow;
}
int dinic() {
    int res = 0, flow;
    while (bfs()) while (flow = find(s, 1e18)) res += flow;
    return res;
}

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    memset(h, -1, sizeof h);
    cin >> n;
    for (int i = 1; i <= n; i ++)
        cin >> a[i];
    s = 0, t = 2 * n + 1;

    for (int i = 1; i <= n; i ++) {
        dp[i] = 1, add(i, i + n, 1);
        std::vector<int> opt;
        for (int j = 1; j < i; j ++)
            if (a[j] <= a[i] && dp[j] + 1 > dp[i])
                opt.clear(), opt.emplace_back(j), dp[i] = dp[j] + 1;
            else if (a[j] <= a[i] && dp[j] + 1 == dp[i])
                opt.emplace_back(j);
        for (auto v : opt)
            add(n + v, i, 1);
    }
    int res = 0;
    for (int i = 1; i <= n; i ++)
        res = max(res, dp[i]);
    cout << res << endl;

    for (int i = 1; i <= n; i ++) {
        if (dp[i] == res)
            add(i + n, t, 1);
        if (dp[i] == 1)
            add(s, i, 1);
    }
    res = dinic();
    cout << res << endl;

    for (int i = 0; i < idx; i += 2) {
        if (e[i ^ 1] == 1 && e[i] == 1 + n || e[i ^ 1] == 1 + n && e[i] == t || e[i ^ 1] == s && e[i] == 1)
            f[i] = 1e18;
        else if (e[i ^ 1] == n && e[i] == n + n || e[i ^ 1] == n + n && e[i] == t || e[i ^ 1] == s && e[i] == n)
            f[i] = 1e18;
    }
    res += dinic();
    cout << min(res, n) << endl;

    return 0;
}

POJ3498 March of the Penguins

参考代码

#include 
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 2e2 + 10, M = 2e4 + 10;

int n, s, t;
double ld;
int h[N], e[M], ne[M], f[M], idx;
int d[N], cur[N];
struct Node {
    int x, y;
    int tot, cnt;
    double operator- (const Node &tmp)const {
        int a = x - tmp.x, b = y - tmp.y;
        return sqrt(a * a * 1.0 + b * b * 1.0);
    }
}pg[N];
void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], f[idx] = c, h[a] = idx ++;
    e[idx] = a, ne[idx] = h[b], f[idx] = 0, h[b] = idx ++;
}
bool bfs() {
    memset(d, -1, sizeof d);
    queue<int> q;
    q.emplace(s), d[s] = 0, cur[s] = h[s];
    while (q.size()) {
        int u = q.front();
        q.pop();

        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (d[j] == -1 && f[i]) {
                d[j] = d[u] + 1, cur[j] = h[j];
                if (j == t) return 1;
                q.emplace(j);
            }
        }
    }
    return 0;
}
int find(int u, int lim) {
    if (u == t) return lim;

    int flow = 0;
    for (int i = cur[u]; ~i && flow < lim; i = ne[i]) {
        cur[u] = i;
        int j = e[i];
        if (d[j] == d[u] + 1 && f[i]) {
            int tmp = find(j, min(lim - flow, f[i]));
            if (!tmp) d[j] = -1;
            f[i] -= tmp, f[i ^ 1] += tmp, flow += tmp;
        }
    }
    return flow;
}
int dinic() {
    int res = 0, flow;
    while (bfs()) while (flow = find(s, 1e18)) res += flow;
    return res;
}

void solve() {
    cin >> n >> ld;

    int sum = 0;
    for (int i = 1; i <= n; i ++)
        cin >> pg[i].x >> pg[i].y >> pg[i].tot >> pg[i].cnt, sum += pg[i].tot;

    s = 0;
    std::vector<int> res;
    for (t = 1; t <= n; t ++) {
        memset(h, -1, sizeof h);
        idx = 0;
        for (int i = 1; i <= n; i ++) {
            add(s, i, pg[i].tot), add(i, i + n, pg[i].cnt);
            for (int j = 1; j <= n; j ++)
                if (i != j && pg[j] - pg[i] <= ld)
                    add(i + n, j, 1e18);
        }
        if (dinic() == sum)
            res.emplace_back(t);
    }

    if (res.empty())
        cout << -1 << endl;
    else {
        for (auto v : res)
            cout << v - 1 << " ";
        cout << endl;
    }
}

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    int dt;

    cin >> dt;

    while (dt --)
        solve();

    return 0;
}