Flow

ianwen

Flow Networks

網路流

# Flow Networks

網路的定義:

有向圖
邊權是容量(capacity) ，c(u, v) > 0
c(u, v) = 0時，edge(u, v)即不存在
一張圖有源點(s) 跟匯點(t)
對於每個點v，必存在 s → v → t

# Flow Networks

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

這是一張網路

# Flow Networks

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

這是一張網路

源點

# Flow Networks

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

這是一張網路

匯點

# Flow Networks

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

這是一張網路

容量

# Flow Networks

Flow(流)的定義:

一個節點到另一個節點的"流動"量
以 f(u, v) 表示，流向為u → v
有正負，代表方向，即: f(u, v) = -f(v, u)
流量不能大於容量

$V_{2}$

$V_{1}$

10/12

f(V1, V2) = 10

f(V2, V1) = -10

# Flow Networks

Flow(流)守恆:

除源點與匯點外，每個節點的總流入=總流出
源點的總流出=匯點總流入

$V_{1}$

10/12

3/6

1/4

x/19

x=12

# Flow Networks

Flow(流)的🌰:

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

11/16

12/12

8/13

8/14

1/9

1/4

4/4

4/7

15/20

f(S, V1) + f(S, V2) = 19

f(V3, T) + f(V4, T) = 19

一樣耶~

Maximum-flow

最大流問題

# Maximum-flow

最大化流量:

給定一張網路與每個邊的容量
求最大有多少流量可以流過這張圖
即求 f(源點, 匯點) 最大值

# Maximum-flow

Dinic's Algorithm 迪尼茨演算法

時間複雜度
運用 Blocking Flow 與 Level Graph

$O(VE^2)$

由 Yefim Dinitz 於1970年提出

# Maxmium-flow

Blocking Flow (阻塞流)

在此情形下，已沒有路徑可以從源點走到匯點
一張圖可以有多種Blocking Flow
Maximum Flow必為Blocking Flow

# Maximum-flow

Blocking Flow (阻塞流)的🌰

$S$

$T$

$V_{1}$

$V_{2}$

2/9

2/2

0/9

$S$

$T$

$V_{1}$

$V_{2}$

9/9

0/2

2/2

2/9

9/9

f(S, T) = 2

f(S, T) = 11

Maximum Flow

Blocking Flow

# Maximum-flow

找 Blocking Flow (阻塞流)

找一條源點走到匯點的路徑
對路徑上的每條邊容量 -= 路徑上最小容量
回到第一步，直到找不到任何一條路徑
得到其中一種Blocking Flow

# Maximum-flow

找 Blocking Flow (阻塞流)

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

# Maximum-flow

找一條路徑

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

# Maximum-flow

減去最小值

$V_{3}$

$V_{1}$

$S$

$T$

$V_{4}$

$V_{2}$

16-12

12-12

20-12

# Maximum-flow

找一條路徑

$S$

$T$

$V_{4}$

$V_{2}$

$V_{3}$

$V_{1}$

# Maximum-flow

減去最小值

$S$

$T$

$V_{4}$

$V_{2}$

13-7

14-7

7-7

$V_{3}$

$V_{1}$

8-7

# Maximum-flow

找一條路徑

$S$

$T$

$V_{4}$

$V_{2}$

$V_{3}$

$V_{1}$

# Maximum-flow

減去最小值

$S$

$T$

$V_{4}$

$V_{2}$

5-4

7-4

4-4

$V_{3}$

$V_{1}$

# Maximum-flow

找不到下一條路徑，找到阻塞流了

$S$

$T$

$V_{4}$

$V_{2}$

$V_{3}$

$V_{1}$

# Maximum-flow

Level Graph (層次圖)

原圖的子圖
將離出發點的距離作為層次
保留所有節點，僅保留由高層級往低層級的邊

# Maximum-flow

找 Level Graph (層次圖)

$V_{2}$

$V_{1}$

$V_{3}$

$V_{4}$

$T$

$S$

# Maximum-flow

找 Level Graph (層次圖) - BFS

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

$S$

$1$

# Maximum-flow

找 Level Graph (層次圖) - BFS

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

$S$

$1$

$2$

# Maximum-flow

找 Level Graph (層次圖) - BFS

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

$S$

$1$

$2$

$3$

# Maximum-flow

Dinic's Algorithm

建立 Residual Graph (剩餘圖)
建立 Residual Graph 的Level Graph
對Level Graph找Blocking Flow
Residual Graph相應邊減去Blocking Flow減的值
回到第一步驟，直到Level Graph找不到Blocking Flow
原圖 (Capacity) - 剩餘圖(Residual) = 流量(Flow)
最後找到的流量必為最大值

# Maximum-flow

原圖

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

初始化

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

建層次圖

Residual Graph

$S$

$1$

Level Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

建層次圖

Residual Graph

Level Graph

$S$

$1$

$2$

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

建層次圖

Residual Graph

Level Graph

$S$

$1$

$2$

$3$

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

建層次圖

Residual Graph

Level Graph

$S$

$1$

$2$

$T$

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

Level Graph找Blocking Flow

Residual Graph

Level Graph

$S$

$1$

$2$

$T$

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

10/10

4/10

4/4

6/8

4/9

10/10

4/10

# Maximum-flow

Residual Graph減去Blocking Flow

Residual Graph

Level Graph

$S$

$1$

$2$

$T$

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

10/10

4/10

4/4

6/8

4/9

10/10

4/10

10-10

10-4

4-4

8-6

9-4

10-10

10-4

# Maximum-flow

更新Residual Graph

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

$1$

$2$

$T$

10/10

4/10

4/4

6/8

4/9

10/10

4/10

# Maximum-flow

更新Residual Graph

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

Level Graph

$S$

$1$

$2$

$T$

10/10

4/10

4/4

6/8

4/9

10/10

4/10

$V_{3}$

$V_{4}$

$1$

$2$

# Maximum-flow

加入反向邊

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

Level Graph

$S$

$1$

$2$

$T$

10/10

4/10

4/4

6/8

4/9

10/10

4/10

$V_{3}$

$V_{4}$

$1$

$2$

# Maximum-flow

刪除舊的Level Graph，進入新循環

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

建立層次圖

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

$4$

$1$

$2$

$3$

# Maximum-flow

Level Graph找Blocking Flow

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

$4$

$1$

$2$

$3$

# Maximum-flow

Level Graph找Blocking Flow

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

$4$

$1$

$2$

5/6

5/5

5/6

$3$

0/6

# Maximum-flow

Residual Graph減去Blocking Flow

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

6-5

5-5

6-5

$V_{3}$

$V_{4}$

6-0

Level Graph

$S$

$4$

$1$

$2$

5/6

5/5

5/6

$3$

0/6

# Maximum-flow

更新Residual Graph

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

$4$

$1$

$2$

5/6

5/5

5/6

$3$

0/6

# Maximum-flow

更新Residual Graph

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

$4$

$1$

$2$

5/6

5/5

5/6

$3$

0/6

# Maximum-flow

加入反向邊

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

$4$

$1$

$2$

5/6

5/5

5/6

$3$

0/6

# Maximum-flow

刪除舊的Level Graph，進入新循環

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

建立層次圖

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

∞

$1$

∞

# Maximum-flow

Level Graph找Blocking Flow

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

∞

$1$

∞

# Maximum-flow

找不到，跳出迴圈

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Level Graph

$S$

∞

$1$

∞

# Maximum-flow

刪除Level Graph與反向邊

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

# Maximum-flow

Capacity - Residual = Flow

Residual Graph

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

Capacity Graph

減

# Maximum-flow

Maximum Flow = 19

$S$

$T$

$V_{1}$

$V_{2}$

$V_{3}$

$V_{4}$

10/10

9/10

0/1

4/4

6/8

9/9

10/10

9/10

5/6

9+10 = 19

# Maximum-flow

扣

const long long INF = 1LL<<60;
 
struct Dinic {  //O(VEE)
    static const int MAXN = 1005;
    struct Edge{
        int u, v;
        long long cap, rest;
    };
 
    int n, m, s, t, d[MAXN], cur[MAXN];
    vector<Edge> edges;
    vector<int> G[MAXN];
 
    void init(){
        edges.clear();
        for ( int i = 0 ; i < n ; i++ ) G[i].clear();
        n = 0;
    }
 
 
    int add_node(){
        return n++;
    }
 
    void add_edge(int u, int v, long long cap){
        edges.push_back( {u, v, cap, cap} );
        edges.push_back( {v, u, 0, 0LL} );
        m = edges.size();
        G[u].push_back(m-2);
        G[v].push_back(m-1);
    }
    
    bool bfs(){
        fill(d,d+n,-1);
        queue<int> que;
        que.push(s); d[s]=0;
        while (!que.empty()){
            int u = que.front(); que.pop();
            for (int ei : G[u]){
                Edge &e = edges[ei];
                if (d[e.v] < 0 && e.rest > 0){
                    d[e.v] = d[u] + 1;
                    que.push(e.v);
                }
            }
        }
        return d[t] >= 0;
    }
 
    long long dfs(int u, long long a){
        if ( u == t || a == 0 ) return a;
        long long flow = 0, f;
        for ( int &i=cur[u]; i < (int)G[u].size() ; i++ ) {
            Edge &e = edges[ G[u][i] ];
            if ( d[u] + 1 != d[e.v] ) continue;
            f = dfs(e.v, min(a, e.rest) );
            if ( f > 0 ) {
                e.rest -= f;
                edges[ G[u][i]^1 ].rest += f;
                flow += f;
                a -= f;
                if ( a == 0 )break;
            }
        }
        return flow;
    }
 
    long long maxflow(int _s, int _t){
        s = _s, t = _t;
        long long flow = 0, mf;
        while ( bfs() ){
            fill(cur,cur+n,0);
            while ( (mf = dfs(s, INF)) ) flow += mf;
        }
        return flow;
    }
};

# Maximum-flow

題

CSES 1694

'\n'

好欸講完了

flow

By wen Ian

flow

a year ago
331

Flow

Flow Networks

Maximum-flow

'\n'

flow

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