Dinic代码 



1.邻接表开四倍




2.在dfs下一层时，应将max_flow取当前所剩流量与容量比较。




3.bfs的能否流入取决于该边是否满流。




代码：



 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 #include "cmath"
 5 #include "queue"
 6 using namespace std;
 7 const int maxn = 205;
 8 const int inf = 0x7fffffff - 5;
 9 int n ,m ,cur[maxn] ,cnt = 1 ,head[maxn] ,dis[maxn];
10 struct node
11 {
12     int pre ,to ,c ,f ,next;
13 }e[maxn*4];
14 
15 void add(int x ,int y, int z)
16 {
17     e[++cnt].pre = x,e[cnt].to = y,e[cnt].c = z,e[cnt].next = head[x],head[x] = cnt;
18 }
19 
20 bool Dinic_bfs()
21 {
22     memset(dis ,0 ,sizeof(dis));
23     dis[1] = 1;
24     queue<int>q;
25     q.push(1);
26     int tt ,t;
27     while(!q.empty())
28     {
29         tt = q.front();
30         q.pop();
31         for(int i=head[tt]; i;i=e[i].next)
32         {
33             t = e[i].to;
34             if(!dis[t]&&e[i].c>e[i].f)
35             {
36                 dis[t] = dis[tt]+1;
37                 q.push(t);
38                 if(t == n)    return true;
39             }
40         }
41     }
42     return false;
43 }
44 
45 long long Dinic_dfs(long long u ,int max_flow)
46 {
47     if(u == n) return max_flow;
48     int tem = max_flow;
49     long long te = 0,ret = 0;
50     for(int i= cur[u]!=0?cur[u]:head[u]; i; i=e[i].next)
51         if(dis[e[i].to] == dis[u]+1)
52         {
53             te = Dinic_dfs(e[i].to ,min(tem ,(e[i].c - e[i].f)));
54             cur[u] = i;
55             e[i].f+=te;
56             e[i^1].f-=te;
57             ret+=te;
58             tem-=te;
59             if(tem == 0) return ret;
60         }
61     return ret;
62 }
63 
64 long long Dinic()
65 {
66     long long sum = 0;
67     while(Dinic_bfs())
68     {
69         memset(cur ,0 ,sizeof(cur));
70         sum+=Dinic_dfs(1 ,inf);
71     }
72     return sum;
73 }
74 
75 int main()
76 {
77     scanf("%d%d" ,&m, &n);
78     int x ,y ,z;
79     for(int i=1; i<=m; i++)
80     {
81         scanf("%d%d%d" ,&x ,&y ,&z);
82         add(x ,y ,z);
83         add(y ,x ,0);
84     }
85     long long ans = Dinic();
86     cout <<ans <<endl;
87 }




 







            

            
posted @ 
2017-11-25 08:51 
Mnirvana 
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