网络流:最大流之EK算法

网络流主要解决三种问题:最大流、最小流和费用流。

最大流算法主要有三种:EK算法、Dinic算法、SAP算法

本篇博客是关于EK算法的。最坏的情况下,EK算法将达到复杂度O(VE2)。

 1 #include <iostream>
 2 #include <queue>
 3 #include<string.h>
 4 using namespace std;
 5 #define arraysize 201
 6 int maxData = 0x7fffffff;
 7 int capacity[arraysize][arraysize]; //记录残留网络的容量
 8 int flow[arraysize];                //标记从源点到当前节点实际还剩多少流量可用
 9 int pre[arraysize];                 //标记在这条路径上当前节点的前驱,同时标记该节点是否在队列中
10 int n,m;
11 queue<int> myqueue;
12 int BFS(int src,int des)
13 {
14     int i,j;
15     while(!myqueue.empty())       //队列清空
16         myqueue.pop();
17     for(i=1;i<m+1;++i)
18     {
19         pre[i]=-1;
20     }
21     pre[src]=0;
22     flow[src]= maxData;
23     myqueue.push(src);
24     while(!myqueue.empty())
25     {
26         int index = myqueue.front();
27         myqueue.pop();
28         if(index == des)            //找到了增广路径
29             break;
30         for(i=1;i<m+1;++i)
31         {
32             if(i!=src && capacity[index][i]>0 && pre[i]==-1)
33             {
34                  pre[i] = index; //记录前驱
35                  flow[i] = min(capacity[index][i],flow[index]);   //关键:迭代的找到增量
36                  myqueue.push(i);
37             }
38         }
39     }
40     if(pre[des]==-1)      //残留图中不再存在增广路径
41         return -1;
42     else
43         return flow[des];
44 }
45 int maxFlow(int src,int des)
46 {
47     int increasement= 0;
48     int sumflow = 0;
49     while((increasement=BFS(src,des))!=-1)
50     {
51          int k = des;          //利用前驱寻找路径
52          while(k!=src)
53          {
54               int last = pre[k];
55               capacity[last][k] -= increasement; //改变正向边的容量
56               capacity[k][last] += increasement; //改变反向边的容量
57               k = last;
58          }
59          sumflow += increasement;
60     }
61     return sumflow;
62 }
63 int main()
64 {
65     int i,j;
66     int start,end,ci;
67     while(cin>>n>>m)
68     {
69         memset(capacity,0,sizeof(capacity));
70         memset(flow,0,sizeof(flow));
71         for(i=0;i<n;++i)
72         {
73             cin>>start>>end>>ci;
74             if(start == end)               //考虑起点终点相同的情况
75                continue;
76             capacity[start][end] +=ci;     //此处注意可能出现多条同一起点终点的情况
77         }
78         cout<<maxFlow(1,m)<<endl;
79     }
80     return 0;
81 }

 

 1 #include <bits/stdc++.h>
 2  
 3 using namespace std;
 4 const int maxn = 1e3;
 5 const int INF = 0x3f3f3f3f;
 6  
 7 int n,m; //n - Vertices  m - edges
 8 int pre[maxn]; //record predecesor and sign if it is visited
 9 int cap[maxn][maxn]; //record the capacity of residual network
10 int flow[maxn]; //record the residual flow from starting vertex to current vertex
11 queue <int> q;
12  
13 int bfs(int st, int ed)
14 {
15   memset(pre,-1,sizeof(pre));
16   while(!q.empty()) q.pop();
17   pre[st] = 0;
18   flow[st] = INF;
19   q.push(st);
20   while(!q.empty())
21   {
22       int t = q.front();
23       q.pop();
24       if(t == ed) break;
25       for(int i = 1; i <= n; i++)
26       {
27           if(pre[i] == -1 && cap[t][i] > 0)
28           {
29               pre[i] = t;
30               flow[i] = min(flow[t],cap[t][i]);
31               q.push(i);
32           }
33       }
34   }
35   if(pre[ed] == -1)  return -1;
36   else              return flow[ed];
37 }
38  
39  
40 int EK(int st, int ed)
41 {
42     int res = 0; //the augmenting flow
43     int sum = 0; //the max_flow
44     while((res = bfs(st,ed)) != -1)//argumenting path
45     {
46         int k = ed;
47         while(k != st)
48         {
49             int f = pre[k];
50             cap[f][k] -= res;
51             cap[k][f] += res;//reversible edge
52             k = f;
53         }
54         sum += res;
55     }
56     return sum;
57 }
58  
59  
60 int main()
61 {
62    int s,t,c;
63    scanf("%d%d",&n,&m);
64    memset(cap,0,sizeof(cap));
65    for(int i = 0; i < m; i++)
66    {
67        scanf("%d%d%d",&s,&t,&c);
68        cap[s][t] = c;
69    }
70    int ans = EK(1,n);
71    printf("%d\n",ans);
72    return 0;
73 }
74 //这是第二个板子,我怕第一个错了

 

posted @ 2019-11-12 22:14  Lovaer  阅读(401)  评论(0编辑  收藏  举报