算法旨在用尽可能简单的思路解决这个问题。理解算法也应该是一个越看越简单的过程，当你看到算法里的一串概念，或者一大坨代码，第一感觉是复杂，此时最好还是从样例入手。通过一个简单的样例，并编程实现，这个过程事实上就能够理解清楚算法里的最重要的思想，之后扩展。对算法的引理或者更复杂的情况。对算法进行改进。最后，再考虑时间和空间复杂度的问题。
了解这个算法是源于在Network Alignment问题中。图论算法用得比較多。而对于alignment。特别是pairwise alignment， 又常常遇到maximum bipartite matching问题，解决问题，是通过Network Flow问题的解法来实现。
一、Network Flow
Network Flow，指的是在从source 到 destination的路径组成一个network, 每条边有一个capacity, 表示从这条边上能通过的最大信息流，而Network Flow问题则要找出从源到目的地能通过的最大流， Maximum Flow. 信息在流动的过程中须要遵循两个原则;
1. 对于每一个节点，流入和流出的信息必须相等。
2.流过每条边的信息不能超过边上的capacity.
最大流问题和minimum cut是等价的，找最大流也就是找minimum cut，minimum cut是例如以下定义的：
我们要在Network上删除一些边。删除掉这些边后，从source 就没有路径到目的地了，我们要找到尽可能少的边，来达到这个目的，这就是minimum cut。
二、 Ford-Fulkerson算法
第一遍读这个算法的时候。不懂，如今读这个算法，认为非常清晰，如今把算法的思路复述一遍。不知道第一次读的人会不会认为easy理解：
1、 构建Residual graph：由于在原network上已经有了capacity, 如今给定这个网络一个流flow的值， 比如边是（u,v）我们计算capacity-f, 同一时候我们也计算（v,u）,值为f(由于capacity为0)，
假设一条边的这个值为正，则保留，否则删除。
2、augmenting path: 通过1得到的就是Residual graph，这个graph上的从source到destination的全部路径都叫做augmenting path.
3、针对每条augmenting path: 改变path上全部边的capacity,改变规则例如以下（以(u,v)为例）：
找到这条path上的最小的capacity, f,
降低u->v的capacity, 添加v->u的capacity.
算法的时间复杂度 O（m+n）f）,f是max-flow.
代码：
// C++ program for implementation of Ford Fulkerson algorithm
#include <iostream>
#include <limits.h>
#include <string.h>
#include <queue>
using namespace std;
// Number of vertices in given graph
#define V 6
/* Returns true if there is a path from source 's' to sink 't' in
residual graph. Also fills parent[] to store the path */
bool bfs(int rGraph[V][V], int s, int t, int parent[])
{
// Create a visited array and mark all vertices as not visited
bool visited[V];
memset(visited, 0, sizeof(visited));
// Create a queue, enqueue source vertex and mark source vertex
// as visited
queue <int> q;
q.push(s);
visited[s] = true;
parent[s] = -1;
// Standard BFS Loop
while (!q.empty())
{
int u = q.front();
q.pop();
for (int v=0; v<V; v++)
{
if (visited[v]==false && rGraph[u][v] > 0)
{
q.push(v);
parent[v] = u;
visited[v] = true;
}
}
}
// If we reached sink in BFS starting from source, then return
// true, else false
return (visited[t] == true);
}
// Returns tne maximum flow from s to t in the given graph
int fordFulkerson(int graph[V][V], int s, int t)
{
int u, v;
// Create a residual graph and fill the residual graph with
// given capacities in the original graph as residual capacities
// in residual graph
int rGraph[V][V]; // Residual graph where rGraph[i][j] indicates
// residual capacity of edge from i to j (if there
// is an edge. If rGraph[i][j] is 0, then there is not)
for (u = 0; u < V; u++)
for (v = 0; v < V; v++)
rGraph[u][v] = graph[u][v];
int parent[V]; // This array is filled by BFS and to store path
int max_flow = 0; // There is no flow initially
// Augment the flow while tere is path from source to sink
while (bfs(rGraph, s, t, parent))
{
// Find minimum residual capacity of the edhes along the
// path filled by BFS. Or we can say find the maximum flow
// through the path found.
int path_flow = INT_MAX;
for (v=t; v!=s; v=parent[v])
{
u = parent[v];
path_flow = min(path_flow, rGraph[u][v]);
}
// update residual capacities of the edges and reverse edges
// along the path
for (v=t; v != s; v=parent[v])
{
u = parent[v];
rGraph[u][v] -= path_flow;
rGraph[v][u] += path_flow;
}
// Add path flow to overall flow
max_flow += path_flow;
}
// Return the overall flow
return max_flow;
}
// Driver program to test above functions
int main()
{
// Let us create a graph shown in the above example
int graph[V][V] = { {0, 16, 13, 0, 0, 0},
{0, 0, 10, 12, 0, 0},
{0, 4, 0, 0, 14, 0},
{0, 0, 9, 0, 0, 20},
{0, 0, 0, 7, 0, 4},
{0, 0, 0, 0, 0, 0}
};
cout << "The maximum possible flow is " << fordFulkerson(graph, 0, 5);
return 0;
}
三、Maximum Bipartite Matching
解决问题就非常easy了。我们先加入上源和目的地节点。如果是任务分配问题，则源能够有边指向全部人。全部任务有边能够指向目的地，我们要找的是人和任务之间的最优匹配。
代码：
// A C++ program to find maximal Bipartite matching.
#include <iostream>
#include <string.h>
using namespace std;
// M is number of applicants and N is number of jobs
#define M 6
#define N 6
// A DFS based recursive function that returns true if a
// matching for vertex u is possible
bool bpm(bool bpGraph[M][N], int u, bool seen[], int matchR[])
{
// Try every job one by one
for (int v = 0; v < N; v++)
{
// If applicant u is interested in job v and v is
// not visited
if (bpGraph[u][v] && !seen[v])
{
seen[v] = true; // Mark v as visited
// If job 'v' is not assigned to an applicant OR
// previously assigned applicant for job v (which is matchR[v])
// has an alternate job available.
// Since v is marked as visited in the above line, matchR[v]
// in the following recursive call will not get job 'v' again
if (matchR[v] < 0 || bpm(bpGraph, matchR[v], seen, matchR))
{
matchR[v] = u;
return true;
}
}
}
return false;
}
// Returns maximum number of matching from M to N
int maxBPM(bool bpGraph[M][N])
{
// An array to keep track of the applicants assigned to
// jobs. The value of matchR[i] is the applicant number
// assigned to job i, the value -1 indicates nobody is
// assigned.
int matchR[N];
// Initially all jobs are available
memset(matchR, -1, sizeof(matchR));
int result = 0; // Count of jobs assigned to applicants
for (int u = 0; u < M; u++)
{
// Mark all jobs as not seen for next applicant.
bool seen[N];
memset(seen, 0, sizeof(seen));
// Find if the applicant 'u' can get a job
if (bpm(bpGraph, u, seen, matchR))
result++;
}
return result;
}
// Driver program to test above functions
int main()
{
// Let us create a bpGraph shown in the above example
bool bpGraph[M][N] = { {0, 1, 1, 0, 0, 0},
{1, 0, 0, 1, 0, 0},
{0, 0, 1, 0, 0, 0},
{0, 0, 1, 1, 0, 0},
{0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 1}
};
cout << "Maximum number of applicants that can get job is "
<< maxBPM(bpGraph);
return 0;
}
四、对于任务分配问题，还有Hungrian算法，这个后面再讲。此算法的时间复杂度和空间复杂度以及改进也能够探讨