最近又复习了下最大流问题，每次看这部分的内容都会有新的收获。可以说最大流问题的资料网上一搜一大把，根本没有必要自己写；但是大部分资料上的专业术语太多了，初学很难理解，至少我当年学这部分的时候前几次就没有看懂。所以我准备备份一点个人的理解。
图-1如图-1所示，在这个运输网络中，源点S和汇点T分别是1,7，各边的容量为C(u,v)。图中红色虚线所示就是一个可行流。标准图示法如图-2所示：
其中p(u,v) / c(u,v)分别表示该边的实际流量与最大容量。
关于最大流
熟悉了什么是网络流，最大流也就很好理解了。就是对于任意的u∈V-{s},使得p(s,u)的和达到最大。上面的运输网络中，最大流如图-3所示：MaxFlow=p(1,2)+p(1,3)=2+1=3。
在介绍最大流问题之前，先介绍几个概念：残余网络，增广路径，反向弧，最大流定理以及求最大流的Ford-Fulkerson方法。
残余网络 增广路径 反向弧
观察下图-4，这种状态下它的残余网络如图-5所示：
也许现在你已经知道什么是残余网络了，对于已经找到一条从S 到T的路径的网络中，只要在这条路径上，把C(u,v)的值更新为C(u,v)-P(u,v)，并且添加反向弧C(v,u)。对应的增广路径Path为残留网络上从S到T的一条简单路径。图-4中1，2，4，7就是一条增广路径，当然还有1，3，4，7。
此外在未做任何操作之前，原始的有向图也是一个残余网络，它仅仅是未做任何更新而已。
最大流定理
如果残留网络上找不到增广路径，则当前流为最大流；反之，如果当前流不为最大流，则一定有增广路径。
Ford-Fulkerson方法
介绍完上面的概念之后，便可以用Ford-Fulkerson方法求最大流了。为什么叫Ford-Fulkerson方法而不是算法，原因在于可以用多种方式实现这一方法，方式并不唯一。下面介绍一种基于广度优先搜索(BFS)来计算增广路径P的算法：Edmonds-Karp算法。
算法流程如下：
设队列Q：存储当前未访问的节点，队首节点出队后，成为已检查的标点；
Path数组：存储当前已访问过的节点的增广路径；
Flow数组：存储一次BFS遍历之后流的可改进量；
Repeat:
Path清空；
源点S进入Path和Q，Path[S]<-0，Flow[S]<-+∞；
While Q非空 and 汇点T未访问 do
Begin
队首顶点u出对；
For每一条从u出发的弧(u,v) do
If v未访问 and 弧(u,v) 的流量可改进；
Then Flow[v]<-min(Flow[u],c[u][v]) and v入队 and Path[v]<-u；
End while
If(汇点T已访问)
Then 从汇点T沿着Path构造残余网络；
Until 汇点T未被访问
应用实例
这是一道最大流的入门题，题目如下：
http://acm.pku.edu.cn/JudgeOnline/problem?id=1273
Description
Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches
so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that
ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
Input
The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
Output
For each case, output a single integer, the maximum rate at which water may emptied from the pond.
Sample Input
5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10
Sample Output
50
#include<iostream>
#include<queue>
using namespace std;
const int maxn = 201;
int map[maxn][maxn],link[maxn],mi[maxn];
queue<int>que;
int n,m;
int bfs(int start)
{
memset(link,0,sizeof(link));
mi[start] = INT_MAX;
while(!que.empty())que.pop();
que.push(start);
while(!que.empty())
{
int u = que.front();
que.pop();
for(int v = 2; v <= n; v++)
{
/*如果该点没有走过并且从cur到i的水渠的流量不为0时*/
if(map[u][v] > 0&& !link[v])
{
mi[v] = mi[u] < map[u][v] ? mi[u] : map[u][v];
link[v] = u;
if(v == n)
return mi[v]; //一次遍历之后的流量增量
que.push(v);
}
}
}
return 0;
}
int cal()
{
int ans = 0,x,per,now;
while((x = bfs(1)) != 0)
{
ans += x;
now = n;
while( now != 1)
{
per = link[now];
map[per][now] -= x;//正向流量减少，
map[now][per] += x;//反向流量增加
now = per;
}
}
return ans;
}
int main()
{
int u,v,cost;
while( cin >> m >> n)
{
memset(map,0,sizeof(map));
while(m--)
{
cin >> u >> v >> cost;
map[u][v] += cost;
}
cout << cal() << endl;
}
return 0;
}