Drainage Ditches(网络流(EK算法))

计算最大流，EK算法模板题。

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <queue>
 4 using namespace std;
 5 const int maxn=520;
 6 const int maxm=100010;
 7 const int INF=1<<28;
 8 int n,m,s,t,cnt,max_flow;
 9 int head[maxn],pre[maxn],a[maxn];
10 
11 struct node
12 {
13     int u;
14     int v;
15     int cap;
16     int next;
17 } edge[maxm];
18 void init()
19 {
20     memset(head,-1,sizeof(head));
21     cnt = 2;
22 }
23 void add(int u,int v,int cap)//用邻接表存储
24 {
25     edge[cnt].u = u;
26     edge[cnt].v = v;
27     edge[cnt].cap = cap;
28     edge[cnt].next = head[u];
29     head[u] = cnt++;
30 }
31 void EK(int s)
32 {
33     queue<int>q;
34     max_flow = 0;
35     for (;;)
36     {
37         memset(a,0,sizeof(a));
38         q.push(s);
39         a[s] = INF;//源点容量为无穷
40         while(!q.empty())
41         {
42             int u = q.front();
43             q.pop();
44             for (int i = head[u]; i!=-1; i=edge[i].next)
45             {
46                 int v = edge[i].v;
47                 if (!a[v] && edge[i].cap > 0)
48                 {
49                     q.push(v);
50                     pre[v] = i;// 记录V的前驱的下标
51                     a[v] = std::min(a[u],edge[i].cap);//更新每个节点的flow
52                 }
53             }
54         }
55         if(a[t]==0)
56             break;
57         int i;
58         for (int u = t; u!=s; u = edge[i].u)//修改增光路
59         {
60             i = pre[u];
61             edge[i].cap -= a[t];//正向的边容量减去残余量
62             edge[i^1].cap += a[t];//反向的边容量(初始为0)加上残余量
63         }
64         max_flow += a[t];
65     }
66 }
67 int main()
68 {
69     while(~scanf("%d%d",&m,&n))
70     {
71         init();
72         while(m--)
73         {
74             int u,v,w;
75             scanf("%d%d%d",&u,&v,&w);
76             add(u,v,w);
77             add(v,u,0);//加反向边
78 
79         }
80         s = 1;
81         t = n;
82         EK(s);
83         printf("%d\n",max_flow);
84     }
85     return 0;
86 }
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posted @ 2013-08-17 23:49  N_ll  阅读(152)  评论(0)  编辑  收藏  举报