USACO 5.4 Telecowmunication(最大流+枚举)

面对最小割之类的题目，完全木想法。。。

枚举+最大流。。复杂度很大了。。。居然很快的就过了。。

 /*
 ID: cuizhe
 LANG: C++
 TASK: telecow
 */
 #include <cstring>
 #include <cstdio>
 #include <queue>
 #include <algorithm>
 using namespace std;
 #define INF 0xffffff
 struct node
 {
     int u,v,w,next;
 } edge[];
 int first[],flag[],dis[];
 int qu[],qv[];
 int t,n,m,ans;
 int sv,ev;
 void CL()
 {
     t = ;
     memset(first,-,sizeof(first));
 }
 void add(int u,int v,int w)
 {
     edge[t].u = u;
     edge[t].v = v;
     edge[t].w = w;
     edge[t].next = first[u];
     first[u] = t ++;

     edge[t].u = v;
     edge[t].v = u;
     edge[t].w = ;
     edge[t].next = first[v];
     first[v] = t ++;
 }
 void build()
 {
     int i;
     CL();
     for(i = ; i < m; i ++)
     {
         add(qu[i]+n,qv[i],INF);
         add(qv[i]+n,qu[i],INF);
     }
     for(i = ; i <= n; i ++)
     {
         if(!flag[i])
         {
             if(i == sv)
                 add(i,i+n,INF);
             else if(i == ev)
                 add(i,i+n,INF);
             else
                 add(i,i+n,);
         }
     }
 }
 int bfs()
 {
     int u,v,i;
     memset(dis,-,sizeof(dis));
     queue<int> que;
     que.push(sv);
     dis[sv] = ;
     while(!que.empty())
     {
         u = que.front();
         que.pop();
         for(i = first[u]; i != -; i = edge[i].next)
         {
             v = edge[i].v;
             if(edge[i].w > &&dis[v] < )
             {
                 dis[v] = dis[u] + ;
                 que.push(v);
             }
         }
     }
     if(dis[ev] > ) return ;
     else return ;
 }
 int dfs(int u,int step)
 {
     int i,a = ,v;
     if (u == ev) return step;
     for (i = first[u]; i != -; i = edge[i].next)
     {
         v = edge[i].v;
         if (edge[i].w > && dis[v] == dis[u]+&&(a = dfs(v,min(step,edge[i].w))))
         {
             edge[i].w -= a;
             edge[i^].w += a;
             return a;
         }
     }
     return ;
 }
 int dinic()
 {
     int res,ans = ;
     while(bfs())
     {
         while(res = dfs(sv,INF))
             ans+= res ;
     }
     return ans;
 }
 void path(int x)
 {
     int i;
     if(x == ) return ;
     for(i = ; i <= n; i ++)
     {
         if(flag[i] == ) continue;
         if(i == sv||i == ev) continue;
         flag[i] = ;
         build();
         if(dinic() == x-)
         {
             if(x == ans)
                 printf("%d",i);
             else
                 printf(" %d",i);
             path(x-);
             return ;
         }
         else
             flag[i] = ;
     }
     return ;
 }
 int main()
 {
     int i;
     freopen("telecow.in","r",stdin);
     freopen("telecow.out","w",stdout);
     scanf("%d%d%d%d",&n,&m,&sv,&ev);
     for(i = ; i < m; i ++)
     {
         scanf("%d%d",&qu[i],&qv[i]);
     }
     build();
     printf("%d\n",ans = dinic());
     path(ans);
     printf("\n");
     return ;
 }