手撕Ford-Fulkerson algorithm 学一半的笔记

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• 定义大概就这些
• 伪代码
• 自己做slide里的quiz
• 搬运别人的代码

我明白了，
余量网络 名如其名
比如你f/c=3/5
那么正边2，reverse edge3,加起来是5

在这个你建的新图上找s到t的路径
然后path的最小边权叫delta

给流图的对应path的每条边e都加流 delta，或者 反边减delta （反边的情况）
得到新的流，重复
直到余量网络没有s到t的路径
(最大流-最小割定理也行)

求最小费用流
找个最大流，
然后

建立个新图
比如这条边的性质 容量，代价= 6，3，跑着2的流
那么正向边标4，3
反向标 2，-3
新图上找 负环(负环的代价和为负)
然后min{a_1}
流加上这个
流调整了
重复
直到新图上找不到 负环(负环的代价和为负)

不断调整f得到最大流
我吐了都，要想手撕个算法首先还得理解一些定义

我找了个课件边看边学

https://www.cs.princeton.edu/~wayne/kleinberg-tardos/pdf/07NetworkFlowI.pdf

定义大概就这些

Def. The bottleneck capacity of an augmenting path P is the minimum residual capacity of any edge in P

伪代码

其中，AUGMENT()如下

自己做slide里的quiz

搬运别人的代码

// Copyright srcmake 2018.
// C++ Example Ford Fulkerson Algorithm

/* Ford Fulkerson Algorithm:
    // 0. Initialize an adjacency matrix to represent our graph.
    // 1. Create the residual graph. (Same as the original graph.)
    // 2. Create an default parent vector for BFS to store the augmenting path. 
    // 3. Keep calling BFS to check for an augmenting path (from the source to the sink...
        // 4. Find the max flow through the path we found.
        // 5. Update the residual capacities of the edges and reverse edges. 
        // 6. Add this path's flow to our total max flow so far.
*/

// The example graph: https://www.srcmake.com/uploads/5/3/9/0/5390645/maxflow_1_orig.jpg

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

//////////////////////////////////////////////////////////////////////////////////////////////
// See the picture here: https://www.srcmake.com/uploads/5/3/9/0/5390645/adjmatrix_1_orig.jpg
vector< vector<int> > FormAdjMatrix()
    {
    // Our adjacency list.
    vector< vector<int> > adjMatrix;
    
    const int n = 6;
    
    // Initialize our matrix to all 0s.
    for(int i = 0; i < n; i++)
        {
        vector<int> row;
        adjMatrix.push_back(row);
        for(int j = 0; j < n; j++)
            {
            adjMatrix[i].push_back(0);
            }
        }
    
    // First number is the vertex, second is the edge capacity.
    adjMatrix[0][1] = 15;
    adjMatrix[0][2] = 12;
    adjMatrix[1][2] = 9;
    adjMatrix[1][3] = 11;
    adjMatrix[2][1] = 5;
    adjMatrix[2][4] = 13;
    adjMatrix[3][2] = 9;
    adjMatrix[3][5] = 25;
    adjMatrix[4][3] = 8;
    adjMatrix[4][5] = 6;
    
    // Our graph is now represented as an adjacency list. 
    return adjMatrix;
    }
//////////////////////////////////////////////////////////////////////////////////////////////


//////////////////////////////////////////////////////////////////////////////////////////////
// A special BFS version that returns true if there's a path from source to sink.
bool BFS(vector< vector<int> > &resAdjMatrix, int &source, int &sink, vector<int> &parent)
    {
    // Create an array for all nodes we visited. Initialized to false.
    int n = resAdjMatrix.size();
    bool visited[n] = { false };
        
    // Create a queue to check each node.
    queue<int> q;
    
    // Push our source into the queue and mark it as visited. It has no parent.
    q.push(source);
    visited[source] = true;
    parent[source] = -1;
        
    // Keep visiting vertices.
    while(q.empty() == false)
        {
        int u = q.front();
        q.pop();
        
        // Check all of u's friends.
        for(int i = 0; i < n; i++)
            {
            int v = i;
            int capacity = resAdjMatrix[u][v];
            
            // We find a neighbor that hasn't been visited, and the capacity is bigger than 0.
            if(visited[v] == false && capacity > 0)
                {
                // Push the neighbor onto the queue, mark it's parent, and mark it as visited.
                q.push(v);
                parent[v] = u;
                visited[v] = true;
                }
            }
        }
        
    // If the sink got visited, then we found a path to it. 
    if(visited[sink] == true) 
        { return true; }
        
    return false;
    }
//////////////////////////////////////////////////////////////////////////////////////////////


//////////////////////////////////////////////////////////////////////////////////////////////
// Use the Ford Fulkerson algorithm. Return the max flow.
int FordFulkerson(vector< vector<int> > &adjMatrix, int &source, int &sink)
    {
    int maxflow = 0;
    
    // 1. Create the residual graph. (Same as the original graph.)
    vector< vector<int> > resAdjMatrix;
    int n = adjMatrix.size();
    for(int i = 0; i < n; i++)
        {
        vector<int> row;
        resAdjMatrix.push_back(row);
        for(int j = 0; j < adjMatrix[i].size(); j++)
            {
            resAdjMatrix[i].push_back(adjMatrix[i][j]);
            }
        }
    
    // 2. Create an empty parent array for BFS to store the augmenting path. 
    vector<int> parent;
    for(int i = 0; i < n; i++)
        {
        parent.push_back(-1);
        }
    
    // 3. Keep calling BFS to check for an augmenting path (from the source to the sink...
    while(BFS(resAdjMatrix, source, sink, parent) == true)
        {
        // 4. Find the max flow through the path we just found.
        int pathflow = 10000007;
        
        // Go through the path we just found. Iterate through the path.
        int v = sink;
        while(v != source)
            {
            int u = parent[v]; // The parent.
            
            // Update the pathflow to this capacity if it's smaller
            int capacity = resAdjMatrix[u][v];
            pathflow = min(pathflow, capacity);
            
            // Setup for the next edge in the path.
            v = u;
            }
        
        // 5. Update the residual capacities of the edges and reverse edges. 
        v = sink;
        while(v != source)
            {
            int u = parent[v]; // The parent.
            
            // Update the capacities.
            
            resAdjMatrix[u][v] -= pathflow;
            resAdjMatrix[v][u] += pathflow;
            
            // Setup for the next edge in the path.
            v = u;
            }
        
        // 6. Add this path's flow to our total max flow so far.
        maxflow += pathflow;
        }
    
    return maxflow;
    }
//////////////////////////////////////////////////////////////////////////////////////////////


//////////////////////////////////////////////////////////////////////////////////////////////    
int main() 
    {
    cout << "Program started.\n";
    
    // Create our adjacency list.
    vector< vector<int> > adjMatrix = FormAdjMatrix();
    
    // Call FordFulkerson to get the max flow from the source to the sink.
    int source = 0;
    int sink = 6;
    
    for(int i = 0; i < 6; i++)
        {
        for(int j = 0; j < 6; j++)
            {
            int source = i;
            int sink = j;
            
            if(i == j) { continue; }
            
            cout << "The max flow from " << source << " to " << sink << " is: ";
            cout << FordFulkerson(adjMatrix, source, sink) << endl;
            }
        cout << endl;
        }
    
    
    cout << "Program ended.\n";
    
    return 0;
    }
//////////////////////////////////////////////////////////////////////////////////////////////

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