最近又複習了下最大流問題，每次看這部分的內容都會有新的收穫。可以說最大流問題的資料網上一搜一大把，根本沒有必要自己寫；但是大部分資料上的專業術語太多了，初學很難理解，至少我當年學這部分的時候前幾次就沒有看懂。所以我準備備份一點個人的理解。
圖-1如圖-1所示，在這個運輸網絡中，源點S和匯點T分別是1,7，各邊的容量爲C(u,v)。圖中紅色虛線所示就是一個可行流。標準圖示法如圖-2所示：
其中p(u,v) / c(u,v)分別表示該邊的實際流量與最大容量。
關於最大流
熟悉了什麼是網絡流，最大流也就很好理解了。就是對於任意的u∈V-{s},使得p(s,u)的和達到最大。上面的運輸網絡中，最大流如圖-3所示：MaxFlow=p(1,2)+p(1,3)=2+1=3。
在介紹最大流問題之前，先介紹幾個概念：殘餘網絡，增廣路徑，反向弧，最大流定理以及求最大流的Ford-Fulkerson方法。
殘餘網絡 增廣路徑 反向弧
觀察下圖-4，這種狀態下它的殘餘網絡如圖-5所示：
也許現在你已經知道什麼是殘餘網絡了，對於已經找到一條從S 到T的路徑的網絡中，只要在這條路徑上，把C(u,v)的值更新爲C(u,v)-P(u,v)，並且添加反向弧C(v,u)。對應的增廣路徑Path爲殘留網絡上從S到T的一條簡單路徑。圖-4中1，2，4，7就是一條增廣路徑，當然還有1，3，4，7。
此外在未做任何操作之前，原始的有向圖也是一個殘餘網絡，它僅僅是未做任何更新而已。
最大流定理
如果殘留網絡上找不到增廣路徑，則當前流爲最大流；反之，如果當前流不爲最大流，則一定有增廣路徑。
Ford-Fulkerson方法
介紹完上面的概念之後，便可以用Ford-Fulkerson方法求最大流了。爲什麼叫Ford-Fulkerson方法而不是算法，原因在於可以用多種方式實現這一方法，方式並不唯一。下面介紹一種基於廣度優先搜索(BFS)來計算增廣路徑P的算法：Edmonds-Karp算法。
算法流程如下：
設隊列Q：存儲當前未訪問的節點，隊首節點出隊後，成爲已檢查的標點；
Path數組：存儲當前已訪問過的節點的增廣路徑；
Flow數組：存儲一次BFS遍歷之後流的可改進量；
Repeat:
Path清空；
源點S進入Path和Q，Path[S]<-0，Flow[S]<-+∞；
While Q非空 and 匯點T未訪問 do
Begin
隊首頂點u出對；
For每一條從u出發的弧(u,v) do
If v未訪問 and 弧(u,v) 的流量可改進；
Then Flow[v]<-min(Flow[u],c[u][v]) and v入隊 and Path[v]<-u；
End while
If(匯點T已訪問)
Then 從匯點T沿着Path構造殘餘網絡；
Until 匯點T未被訪問
應用實例
這是一道最大流的入門題，題目如下：
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++)
{
/*如果該點沒有走過並且從cur到i的水渠的流量不爲0時*/
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]; //一次遍歷之後的流量增量
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;//正向流量減少，
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;
}
